HANCOCK'S  APPLIED   MECHANICS 
FOR  ENGINEERS 


THE  MACMILLAN  COMPANY 

NEW  YORK   •    BOSTON   •    CHICAGO  -    DALLAS 
ATLANTA  •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  LIMITED 

LONDON  •    BOMBAY   •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  LTD. 

TORONTO 


HANCOCK'S  APPLIED  MECHANICS 
FOR  ENGINEERS 


REVISED   AND   REWRITTEN 
BY 

N.   C.    RIGGS 

R    OF    THEORETICAL   AND   APPLIED    MECHANICS    IN    THB 
SCHOOL   OF   APPLIED   SCIENCE    OF    THE    CARNEGIE 
INSTITUTE   OF   TECHNOLOGY 


THE   MACMILLAN   COMPANY 
1920 

All  right*  referred 


TA35D 


Engineering 

Library 

* 


Engineering 
Jjibrary 


OOPTEIOHT,    1909   AND   1915, 

BY  THE  MACMILLAN  COMPANY. 
Set  up  and  electrotyped.    Published  January,  1909. 


J.  S.  Cushiiig  Co.  — Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE 

IN  the  preparation  of  this  book  the  author  has  had  in 
mind  the  fact  that  the  student  finds  much  difficulty  in 
seeing  the  applications  of  theory  to  practical  problems. 
For  this  reason  each  new  principle  developed  is  followed 
by  a  number  of  applications.  In  many  cases  these  are 
illustrated,  and  they  all  deal  with  matters  that  directly 
concern  the  engineer.  It  is  believed  that  problems  in 
mechanics  should  be  practical  engineering  work.  The 
author  has  endeavored  to  follow  out  this  idea  in  writing 
the  present  volume.  Accordingly,  the  title  "Applied 
Mechanics  for  Engineers  "  has  been  given  to  the  book. 

The  book  is  intended  as  a  text-book  for  engineering 
students  of  the  Junior  year.  The  subject-matter  is  such 
as  is  usually  covered  by  the  work  of  one  semester.  In 
some  chapters  more  material  is  presented  than  can  be  used 
in  this  time.  With  this  idea  in  mind,  the  articles  in 
these  chapters  have  been  arranged  so  that  those  coming 
last  may  be  omitted  without  affecting  the  continuity  of 
the  work.  The  book  contains  more  problems  than  can 
usually  be  given  in  any  one  semester. 

While  it  is  difficult  to  present  new  material  in  the 
matter  of  principles,  much  that  is  new  has  been  intro- 
duced in  the  applications  of  these  principles.  The  sub- 
ject of  Couples  is  treated  by  representing  the  couples 
by  means  of  vectors.  The  author  claims  that  the  chap- 
ters on  Moment  of  Inertia,  Center  of  Gravity,  Work 
and  Energy,  Friction  and  Impact  are  more  complete 
in  theory  and  applications  than  those  of  any  othei 

789552 


VI  PREFACE 

American  text-book  on  the  same  subject.  These  are 
matters  upon  which  the  engineer  frequently  needs  infor- 
mation ;  frequent  reference  is,  therefore,  given  to  origi- 
nal sources  of  information.  It  is  hoped  that  these  chapters 
will  be  especially  helpful  to  engineers  as  well  as  to  students 
in  college,  and  that  they  will  receive  much  benefit  as 
a  result  of  looking  up  the  references  cited.  In  general, 
the  answers  to  the  problems  have  been  omitted  for  the 
reason  that  students  who  are  prepared  to  use  this  book 
should  be  taught  to  check  their  results  and  work  inde- 
pendently of  any  printed  answer. 

The  author  wishes  to  acknowledge  the  helpful  sugges- 
tions obtained  from  the  many  standard  works  on  me- 
chanics. An  attempt  has  been  made  to  give  the  specific 
reference  to  the  original  for  material  taken  from  engi- 
neering works  or  periodical  literature.  He  wishes,  more- 
over, to  express  his  thanks  to  Dean  C.  H.  Benjamin  and 
Professor  L.  V.  Ludy  for  their  careful  reading  of  the 
manuscript,  to  Professor  W.  K.  Hatt  for  many  of  the 
problems  used,  and  to  Dean  W.  F.  M.  Goss,  whose  con- 
tinued interest  and  advice  have  been  a  constant  source  of 
inspiration.  It  is  hoped  that  the  work  may  be  an  inspira- 
tion to  students  of  engineering. 

E.  L.  HANCOCK. 
PURDUE  UNIVERSITY, 
November,  1908. 


PREFACE   TO  THE   REVISED    EDITION 

IN  the  revision  of  this  text,  although  rather  extensive 
changes  in  method  of  treatment  have  been  made  in  cer- 
tain parts,  the  general  subject  matter  and  order  of  arrange- 
ment have  been,  in  the  main,  retained.  The  chapter  on 
Dynamics  of  Machinery,  now  called  Dynamics  of  a  Rigid 
Body,  has  been  transferred  to  a  later  position  in  the  book 
on  account  of  its  relatively  greater  difficulty. 

An  important  change  is  to  be  found  in  the  much  larger 
use  of  graphical  methods.  The  graphical  and  analytical 
methods  have  been  developed  simultaneously  and  many 
problems  are  given  to  be  solved  by  both  methods.  The 
use  of  graphical  methods  has  brought  about  the  intro- 
duction of  considerable  new  material,  particularly  in  the 
construction  of  stress  diagrams  for  trusses,  and  in  the 
application  of  the  equilibrium  polygon  to  centers  of  grav- 
ity of  plane  areas,  to  weighted  strings  and  linkages,  and 
to  bending  moment  diagrams. 

The  problems  illustrating  the  principles  follow  imme- 
diately after  the  development  of  the  principles  but  are  of 
such  a  nature  as  to  require  original  thinking.  The  solu- 
tion does  not  consist  in  finding  and  substituting  in  the 
proper  formula.  About  two  hundred  new  problems  have 
been  added  throughout  the  book. 

I  wish  to  express  my  thanks  to  Mr.  E.  G.  Frazer  for  a 
careful  reading  of  the  proof  and  for  helpful  suggestions. 

N.  C.  RIGGS. 
SCHOOL  OF  APPLIED  SCIENCE, 
CARNEGIE  INSTITUTE  OF  TECHNOLOGY. 
January,  1915. 

vii 


TABLE   OF  CONTENTS 

CHAPTER  I 
ARTICLES  1-16 

PAGK 

DEFINITIONS 1 

Force  —  Inertia  —  Mass  —  Units  —  Vectors  —  Displace- 
ment —  Transmissibility  of  Forces — Force  Triangle  —  Force 
Polygon. 

CHAPTER  II 
ARTICLES  17-24 

CONCURRENT  FORCES 10 

Conditions  for  Equilibrium  —  Equilibrium  of  Three 
Forces  —  Concurrent  Forces  in  Space  —  Moments — Vari- 
gnon's  Theorem. 

CHAPTER  III 
ARTICLES  25-32 

PARALLEL  FORCES 27 

Moment  of  Resultant  of  Parallel  Forces  —  Center  of  Par- 
allel Forces. 

CHAPTER  IV 
ARTICLES  33-43 

CENTER  OF  GRAVITY   .        .        .        .     .  *  -.••>»      ,-< "      .        .      38 

Center  of  Gravity  by  Integration  —  Graphical  Methods  — 
Simpson's  Rule  —  Durand's  Rule  —  Theorem  of  Pappus. 


X  TABLE  OF  CONTENTS 

CHAPTER  V 
ARTICLES  44-53 

PAGB 

COUPLES .        f      71 

Definition  —  Moment  of  a  Couple  —  Combination  of 
Couples  —  Equivalent  Couples  —  Vector  Representation  of 
Couples. 

CHAPTER  VI 
ARTICLES  54-59 

NON-CONCURRENT  FORCES 82 

Forces  in  a  Plane  —  Conditions  of  Equilibrium  —  Forces  in 
Space  —  Graphical  Method  for  Forces  in  One  Plane  —  Equi- 
librium Polygon  —  Stresses  in  Frames  —  Bow's  Notation. 

CHAPTER  VII 

ARTICLES  60-84 

MOMENT  OF  INERTIA 106 

Definition  of  Moment  of  Inertia  —  Radius  of  Gyration  — 
Parallel  Axes  —  Inclined  Axes  —  Product  of  Inertia  —  Prin- 
cipal Axes  —  Graphical  Method  —  Use  of  Simpson's  Rule  — 
Ellipse  of  Inertia  —  Ellipsoid  of  Inertia. 

CHAPTER  VIII 
ARTICLES  85-101 

FLEXIBLE.  CORDS 148 

Introduction  —  Suspended  Weights  —  Graphical  Methods 
—  String  Polygon  —  The  Linked  Arch  —  Masonry  Arch  — 
Bending  Moments — Tensile  Stresses  in  Beams — Shear  — 
Cords  and  Pulleys  —  Weighted  Cord  as  Parabola  —  Cate- 
nary—  Hyperbolic  Functions. 


TABLE  OF  CONTENTS 


CHAPTER  IX 
ARTICLES  102-114 

PAQB 

MOTION  IN  A  STRAIGHT  LINE     .        .        .        .-•*.«.    183 

Velocity  —  Acceleration  —  Newton's  Laws  of  Motion  — 
Harmonic  Motion  —  Motion  under  Different  Laws  of  Force 
—  Relative  Velocity. 


CHAPTER  X 
ARTICLES  115-124 

CURVILINEAR  MOTION         .        .        .        •        '•        .        .        .    204 

Velocity  —  Acceleration  —  Centrifugal  Force  —  Motion  in 
a  Vertical  Plane  —  Simple  Pendulum  —  Cycloidal  Pendulum 
—  Projectiles  — Motion  in  a  Twisted  Curve. 


CHAPTER  XI 
ARTICLES  125-133 

ROTARY  MOTION  ....        .    ,,*        •        .        .        .    232 

Angular  Velocity  and  Angular  Acceleration  —  Plane  Mo- 
tion of  a  Body  —  Instantaneous  Axis  of  Rotation  —  Combi- 
nation of  Simultaneous  Rotations  —  Foucault's  Pendulum. 


CHAPTER  XII 
ARTICLES  134-153 

WORK  AND  ENERGY     .        . 245 

Definitions  —  Friction  Forces  —  Graphical  Representation 
of  Work  —  Conservation  of  Energy  —  Pile  Driver  —  Steam 
Hammer  —  Work  of  Impressed  Forces  —  Brake-shoe  Test- 
ing Machine  —  Kinetic  Energy  of  Rolling  Bodies. 


Xll  TABLE  OF  CONTENTS 

CHAPTER  XIII 
ARTICLES  154-176 

PAGE 

FRICTION v       .        .    283 

Coefficient  of  Friction  —  Laws  of  Friction  for  Dry  Sur- 
faces —  Friction  of  Lubricated  Surfaces  —  The  Wedge  — 
Method  of  Testing  Lubricants  —  Rolling  Friction  —  Anti- 
friction Wheels  —  Resistance  of  Roads  —  Roller  Bearings  — 
Ball  Bearings  —  Friction  Gears  —  Friction  of  Belts  —  Power 
Transmitted  by  a  Belt — Transmission  Dynamometer  — 
Creeping  of  Belts  —  Friction  of  a  Worn  Bearing  —  Friction 
of  Pivots  —  Absorption  Dynamometer  —  Friction  Brake  — 
Prony  Friction  Brake  —  Friction  of  Brake  Shoes  —  Train 
Resistance. 

CHAPTER  XIV 
ARTICLES  177-206 

DYNAMICS  OF  RIGID  BODIES 334 

D'Alembert's  Principle — Translation  of  a  Rigid  Body  — 
Rotation  about  a  Fixed  Axis  —  Equations  of  Motion  of  a 
Rotating  Body  —  Reactions  of  Supports  of  a  Rotating  Body 

—  Compound  Pendulum  —  Centers  of  Oscillation  and  Sus- 
pension—  Experimental  Determination  of  Moment  of  In- 
ertia —  Determination    of    g  —  The    Torsion    Balance  — 
Rotating  Body  under  the  Action  of  No  Forces  —  Rotation 
of  a  Body  about   a   Fixed   Axis  with    Constant  Angular 
Velocity  —  Rotation  of  Locomotive  Drive  Wheel  —  Stand- 
ing and  Running  Balance  of  a  Shaft  —  Rotation  of  Fly- 
wheel of  Steam  Engine  —  Rotation  and  Translation  —  The 
Connecting  Rod  —  Angular  Momentum  —  Torque  and  An- 
gular Momentum  —  Moment  of  Momentum  of  a  Body  with 
One  Point  Fixed  —  Vector  Representation  of  Angular  Mo- 
mentum —  Kinetic  Energy  of  a  Body  with  One  Fixed  Point 

—  Vector  Rate  of  Change  Due  to  Rotation  —  Rate  of  Change 
of  Angular  Momentum  of  a  Body  Due  to  Rotating  Axes  — 


TABLE  OF  CONTENTS  xiii 

PA.GB 

Motion  of  Center  of  Gravity  of  a  Body  —  Gyroscope,  Hori- 
zontal Axis—  The  Gyroscopic  Couple  —  Car  on  Single  Rail 

—  Gyroscope,  Inclined  Axis. 

CHAPTER  XV 

ARTICLES  207-215 

IMPACT 390 

Definitions  —  Direct  Central  Impact  —  Momentum  and 
Kinetic  Energy  in  Impact  —  Elasticity  of  Material  —  Im- 
pact Tension  and  Impact  Compression  —  Direct  Eccentric 
Impact  —  Centers  of  Percussion  and  Instantaneous  Rotation 

—  Oblique  Impact  of  Body  against  Smooth  Plane  —  Impact 
of  Rotating  Body. 


APPLIED  MECHANICS  FOR  ENGINEERS 

CHAPTER  I 
DEFINITIONS 

1.  Introduction.  —  The  study  of  the  subject  of  mechanics 
involves  a  study  of  matter,  space,  and  time,  and  of  the  be- 
havior of  bodies  under  the  action  of  forces.     The  subject 
as   presented  in   this   book   consists  of  two  parts  ;    viz. 
statics,  including  the  study  of  bodies  under  the   action 
of  systems  of  forces  that  are  in  equilibrium  (balanced), 
and  dynamics,  including  a  study  of  the  motion  of  bodies. 

2.  Force.  —  A   body   acted   upon   by  the  attraction   or 
repulsion  of  another  body  is  said  to  be  subjected  to  an 
attractive  or  repulsive  force,  as  the  case  may  be.     In  sim- 
ple terms  a  force  is  a  push  or  a  pull.     Forces  are  usually 
defined  by  the  effects  produced  by  them,  as,  for  example, 
we  say,  a  force  is  something  that  produces  motion   or 
tends  to  produce  motion,  or  changes  or  tends  to  change 
motion,   or   that   changes  the  size  or  shape  of  a  body. 
Forces  always  occur  in  pairs;  for  example,  a  book  held 
in  the  outstretched  hand  exerts  a  downward  pressure  on 
the  hand,  and  the  hand  exerts  an  equal  upward  pressure 
on  the  book. 

3.  Inertia.    Mass.  —  It  is  a  matter  of  common  experience 
that  bodies  vary  in  the  amount  of  resistance  that  they 
offer  to  a  change  of  their  state  of  rest  or  motion.     Thus 

B  1 


2  'APPtiEI)  MECHANICS  FOR  ENGINEERS 

it  is  more  difficult  to  stop  a  swiftly  moving  ball  of  iron 
than  one  of  wood  having  the  same  dimensions  and  speed. 
It  is  easier  to  set  in  motion  the  wooden  ball  than  to  give 
the  same  speed  to  the  iron  ball  of  the  same  size. 

The  property  of  resistance  to  change  of  its  state  of  rest 
or  motion  that  every  body  has  is  called  its  inertia. 

The  word  mass  is  used  in  the  sense  of  a  measure  of  inertia. 

4.  The   Unit  of   Mass.  — A   certain   arbitrary   piece   of 
platinum  carefully  preserved  by  the  British  government 
is  known  as  the  standard  mass  of  one  pound  avoirdupois. 
Any  other  piece  of  matter  which  under  the  action  of  a 
certain  force  has  its  motion  changed  in  the  same  way  as 
the  standard  mass  of  one  pound  under  the  same  condi- 
tions is  also  called  a  mass  of  one  pound. 

The  corresponding  unit  in  the  French  system  is  the  gram. 

5.  The  Unit  of  Force.  —  The  pull,  or  attraction,  that  the 
earth  exerts  upon  a  mass  of  one  pound  at  sea  level  and 
latitude  45°  is  called  the  force  of  one  pound.     Similarly, 
the  force  of  one  gram  is  the  force  that  the  earth  exerts 
upon  a  mass  of  one  gram  at  sea  level  and  latitude  45°. 
These  units  of  force  are  called  the  gravitational  units  of  force. 

The  poundal  is  defined  as  that  force  which  acting  alone 
on  a  mass  of  one  pound  for  one  second  produces  in  that 
mass  a  change  of  velocity  of  one  foot  per  second. 

The  dyne  is  defined  as  that  force  which  acting  alone  on 
a  mass  of  one  gram  for  one  second  produces  in  that  mass 
a  change  of  velocity  of  one  centimeter  per  second. 

The  poundal  and  dyne  are  called  absolute  units  of  force. 

The  force  with  which  the  earth  attracts  a  body  is  called 


DEFINITIONS 


the  weight  of  the  body.  Mass  differs  from  weight,  in  that 
the  weight  varies  with  the  position  on  the  surface  of  the 
earth  and  with  the  height  above  the  surface,  while  the 
mass  remains  the  same.  The  weight  of  a  body  may  be 
determined  by  means  of  the  spring  balance.  The  equal- 
armed  balance  gives  the  same  weight  regardless  of  dis- 
tance from  the  center  of  the  earth.  The  equal-armed 
balance  really  measures  mass. 

6.  Unit  Weight.  —  The  weight  of  a  cubic  foot  of  a  sub- 
stance will  be  called  the  unit  weight  of  the  substance  and 
will  be  represented  by  7.  Below  is  given  a  table  of  such 

TABLE  I 

UNIT  WEIGHTS  AND  SPECIFIC  GRAVITY  OF  SOME  MATERIALS 
(Kent's  "Engineer's  Pocket  Book") 


MATERIAL 

SPECIFIC  GRAVITY 

UNIT  WEIGHT 

Brick 

Soft 

1.6 

100 

Hard 

2.0 

125 

Fire 

2.24-2.4 

140-150 

Brickwork  —  mortar 

1.6 

100 

Brickwork  —  cement 

1.79 

112 

Concrete 

1.92-2.24 

120-140 

Copper 
Earth  —  loose 

8.85 
1.15-1.28 

552 

72-80 

Earth  —  rammed 

1.44-1.76 

90-110 

Iron  —  cast 

7.21 

450 

Iron  —  wrought 
Masonry  —  dressed 
Pine  —  white 

7.7 
2.24-2.88 
.45 

480 
140-180 

28 

Pine  —  yellow 
Steel 

.61 

7.85 

38 
490 

White  Oak 

.77 

48 

4  APPLIED  MECHANICS  FOR  ENGINEERS 

weights  at  sea  level.  The  unit  weight  of  a  substance 
divided  by  the  unit  weight  of  pure  water  gives  its  specific 
gravity. 

7.  Rigid  Body.  —  In  studying  the  state  of  motion  or 
rest  of  a  body  due  to  the  application  of  forces  acting 
upon  it,  the  deformation  of  the  body  itself,  due  to  the 
forces,  may  be  disregarded.      When  so  considered,  it  is 
customary  to  say  that  the  body  is  a  rigid  body.     Unless 
otherwise  stated  bodies  will  be  considered  as  rigid  bodies 
in  this  book. 

8.  Vectors.  —  Any   quantity  that   has    magnitude    and 
direction  may  be  represented  graphically  by  a  directed 
segment  of  a  straight  line.     The  length  of  the  segment 
is  taken  to  measure  the  magnitude  of  the  quantity,  and 
the  direction  of  the  segment  to  indicate  the  direction  of 
the  quantity. 

The  directed  segment  of  the  line  is  called  a  vector,  and 
the  quantity  it  represents,  a  vector  quantity. 

9.  Displacement. — By  the  displacement  of   a  body  is 
meant  its  change  from  one  position  to  another. 

Since  a  change  of  position  involves  magnitude  and  direc- 
tion, a  displacement  may  be  represented  by  a  vector,  the 
length  of  the  vector  representing  the  distance  from  the  first 
position  to  the  second  and  the  direction  of  the  vector  repre- 
senting the  direction  of  the  second  position  from  the  first. 

From  the  definition  of  a  displacement  it  follows  that 
two  successive  displacements  are  equivalent  to  a  single 
displacement.  Thus,  if  a  man  walks  due  east  one  mile 
and  then  due  north  one  mile,  we  might  represent  his  dis- 


DEFINITIONS  5 

placement  from  the  original  position  by  a  vector  drawn 
northeast  to  represent  a  length  equal  to  V2"  miles.  Or, 
in  Fig.  1,  if  Pl  represents  a  displacement  of  a  body  in 
the  direction  indicated  and  P2  a  subsequent  displacement 
in  the  direction  of  P^  then  R  represents  a  displacement 
equivalent  to  Pl  and  P2.  It  is  seen  that  R  may  be  deter- 
mined by  constructing  a  parallelogram  on  P1  and  P2  as 
sides  and  drawing  the  diagonal. 


FIG.  1 

10.  Representation  of  Forces.  —  A  force  has  a  certain 
magnitude,  acts  in  a  certain  direction,  and  has  a  definite 
point  of  application.  If  a  man,  for  example,  attaches  a 
rope  to  a  log  and  pulls  on  the  rope,  his  pull  may  be  meas- 
ured in  pounds,  it  acts  along  the  rope,  and  its  point  of 
application  is  the  point  of  attachment  of  the  rope  to  the 
log.  It  is  convenient,  for  the  purpose  of  analysis,  to 
represent  forces  by  vectors,  the  length  of  the  vector 
representing  the  magnitude  of  the  force  and  its  direction 
giving  the  direction  in  which  the  force  acts.  Thus,  a 
10-lb.  force,  acting  in  a  direction  30°  with  the  horizontal,  is 
represented  by  a  vector  drawn  in  the  same  direction  and 
having  its  point  of  application  in  the  body,  and  having  a 
length  representing  10  Ib.  (In  this  case,  if  1  in.  represents 
2  Ib.,  the  length  of  the  vector  is  5  in.)  The  line  along 
which  a  force  acts  will  be  referred  to  as  its  line  of  action. 


6  APPLIED  MECHANICS  FOB  ENGINEERS 

11.  Transmissibility  of  Forces.  —  It  is  a  matter  of  experi- 
ence that  the  point  of  application  of  a  force  may  be  changed 
to  any  point  along  its  line  of  action  without  changing  the 
effect  of  the  force  upon  the  rigid  body.    This,  of  course,  is  on 
the  assumption  that  all  the  force  is  transmitted  to  the  body. 
The  law  may  be  stated  as  follows  :   The  point  of  application 
of  a  force  may  be  transferred  anywhere  along  its  line  of  action 
without  changing  its  effect  on  any  rigid  body  on  which  it  acts. 

12.  Concurrent  Forces.  —  When  the  lines  of  action  of 
two   or   more   forces   pass   through  the  same  point,  the 
forces  are  known  as  concurrent  forces. 

13.  Resultant  of  Two  Concurrent  Forces.  —  If    two  con- 
current forces  act  on  a  body,  there  is  some  single  force 
that  might  be  applied  at  the  point  of  intersection  of  the 
forces  to  produce  the  same  effect.     This  single  force  is 
called  the  resultant  of  the  two  forces.     Experiment  shows 
that  it  may  be  found  as  follows :  construct  upon  the  vec- 
tors representing  the  forces,  laid  off  from  the  intersection 
of  their  lines  of   action,  a  parallelogram  and  draw  the 
diagonal  from  the  point   of    application.     This  diagonal 
represents  the  resultant  of  the  two  forces  in  magnitude 
and  direction  and  it  is  the  line  of  action  of  the  resultant. 


o 

FIG.  2 


Thus,  if  Pl  and  P2  (Fig.  2)  are  the  forces,  then  E  is 
the  resultant. 


DEFINITIONS 


Algebraically  JB  =  VJV  +  IV  +  2  IM>a  cos 

Instead  of  speaking  of  the  vector  which  represents  a 
force,  we  shall  for  the  sake  of  brevity  speak  of  the  vector 
as  the  force. 

14.  Resolution  of  Force.  —  We  have  just  seen  how  two 
concurrent  forces  may  be  replaced  by  a  single  force  called 
their  resultant.     In  a  similar  way  a  single  force  may  be 
resolved  into  two  forces.     These  forces  are  the  sides  of  a 
parallelogram  of  which  the  single  force  is  a  diagonal.     It 
is  clear,  then,  that  there  are  an  infinite  number  of  compo- 
nents into  which  a  single  resultant  may  be  resolved.     It 
is  necessary,  therefore,  in  speaking  of  the  components  of 
a  force,  to  state  specifically  which  are  intended.     It  will 
be  seen  in  problems  that  follow  that  the  components  most 
often  used   are    at   right  angles  to  each  other,   and  are 
.usually  the  vertical  and  horizontal  components.     In  such  a 
case  the  components  are  the  projections  of  the  force  on  the 
vertical  and  horizontal  lines. 

15.  Force  Triangle.  —  It  follows  directly  from  the  paral- 
lelogram law  of  forces  that  if  we  draw  from  any  point  a  line 
parallel  and  equal  to  one  of  two  concurrent  forces,  P2  say, 
and  from  the  extremity  of  this  line  another  line  parallel 
and  equal  to  Pv  then  the  remaining  side  of  the  triangle 
will  represent  the  resultant   R.     This  triangle  is  called 
the  force  triangle.     In  general,  the  resultant  of  two  con- 
current forces  may  be  found  by  drawing  lines  parallel  to 
the  forces  as  above.     The  line  necessary  to  complete  the 
triangle  is  the  resultant,  and  its  direction  is  always  away 
from  the  point  of  application.     The  equal  and  opposite 


8 


APPLIED  MECHANICS  FOR   ENGINEERS 


of  this  resultant  would  be  a  single  force  that  would  hold 
the  two  concurrent  forces  in  equilibrium. 

The  single  force  which  will  balance  a  given  set  of  forces 
is  called  their  equilibrant. 

16.  Force  Polygon.  —  If  more  than  two  forces  are  con- 
current, we  may  find  their  resultant  by  proceeding  in  a 
way  similar  to  that  outlined  above.  Thus,  let  the  forces 


be  P 


P3,  P4,  etc.   (Fig.  3),  all  passing  through  a 


FIG.  3 

point ;  from  any  point  draw  a  line  equal  and  parallel  to  Pv 
from  the  extremity  of  the  line  draw  another  equal  and 
parallel  to  P2,  from  the  extremity  of  this  last  line  draw 
another  equal  and  parallel  to  P3,  and  proceed  in  the  same 
way  for  the  other  forces.  The  figure  produced  will  be  a 
polygon  whose  sides  are  equal  and  parallel  to  the  forces. 
The  resultant  of  the  given  forces  is  then  represented  in 
magnitude  and  direction  by  the  line  necessary  to  close  the 
polygon,  and  its  line  of  action  passes  through  the  intersec- 
tion of  the  given  concurrent  forces. 

The  arrow,  representing  the  direction  of  the  resultant, 
will  always  be  away  from  the  point  of  application.  (See 
Fig.  3.) 

By  drawing  the  lines   OA,   OB,  00,  etc.,  it  is  easy  to 


DEFINITIONS  9 

see  that  OA  represents  the  resultant  of  Pl  and  P2,  that 
OB  represents  the  resultant  of  OA  and  P3,  and  so  of  Pv 
P2,  and  P3,  etc.  That  is,  the  force  polygon  follows 
directly  from  the  force  triangle.  If  the  polygon  be  closed, 
the  system  of  forces  will  be  in  equilibrium,  and  con- 
versely. The  single  force  necessary  to  produce  equi- 
librium will,  in  any  case,  be  equal  and  opposite  to  the 
resultant  R.  The  student  should  construct  force  polygons 
by  taking  the  forces  in  different  orders  and  checking  the 
resultant  in  each  case. 

By  means  of  the  force  polygon  it  is  easy  to  find  graph- 
ically the  resultant  of  any  number  of  concurrent  forces 
in  a  plane.  The  work,  however,  must  be  done  accurately. 

The  student  should  show  that  the  force  polygon  may 
be  used  for  finding  the  resultant  of  concurring  forces  in 
space,  by  considering  two  forces  at  a  time.  The  force 
polygon  in  this  case  is  called  a  twisted  polygon. 

Problem  1.  Find  graphically  the  resultant  in  magnitude,  direc- 
tion, and  point  of  application  of  the  following  four  concurrent  forces 
in  one  plane :  80  Ib.  in  direction  E.,  60  Ib.  E.  50°  N.,  100  Ib.  W. 
40°  N.,  and  120  Ib.  E.  30°  S.  Check  by  taking  the  forces  in  different 
orders. 

Problem  2.  Find  graphically  the  resultant  of  the  following  four 
concurrent  forces :  (a)  50  Ib.  directed  E.  in  a  horizontal  plane. 
(6)  80  Ib.  directed  N.  40°  W.  in  the  horizontal  plane,  (c)  100  Ib. 
directed  30°  above  the  horizontal,  the  projection  of  the  force  on  the 
horizontal  plane  being  directed  S.  20°  E.,  and  (d)  70  Ib.  directed  60° 
above  the  horizontal,  the  projection  of  the  force  on  the  horizontal 
plane  being  directed  W.  40°  S. 

SUGGESTION.  Resolve  the  forces  (c)  and  (ef)  into  their  horizontal 
and  vertical  components  before  combining  with  the  other  forces. 


CHAPTER   II 


CONCURRENT  FORCES 

17.  Concurrent  Forces  in  a  Plane.  —  It  will  often  be 
convenient  to  consider  forces  as  acting  on  a  material 
point ;  this  is  equivalent  to  considering  the  mass  of  the 

body  as  concen- 
trated at  a  point. 
Given  a  material 
point  0  (Fig.  4) 
acted  upon  by  a 
number  of  forces 
in  a  plane,  Pv 
PV  P3,  P4,  etc., 
making  angles 


respectively, 
with  the  posi- 
tive a>axis,  it  is 
desired  to  find 

*  the  resultant  of 

all  of  them  in  magnitude  and  direction  ;  that  is,  the  single 
ideal  force  that  would  produce  the  same  effect  as  the  sys- 
tem of  forces. 

Each  force  P  may  be  resolved  into  components  along 
the  x-  and  y-axes,  giving  P  cos  a  along  the  #-axis,  and 

10 


CONCURRENT  FORCES 


11 


P  sin  a  along  the  «/-axis.     The  sum  of  these  components 
along  the  #-axis  may  be  expressed, 

^X—  Pl  cos  «!  +  P2  cos  02  +  ^3  cos  «3  +  etc., 
the  proper  algebraic  sign  being  given  cos  a  in  each  case. 
In  a  similar  way  the  sum  of  the  components  along  the 
?/-axis  may  be  written, 

2  Y=  P1  sin  «!  +  P2  sin  a^  +  PB  sin  «3  +  etc. 

These  forces,  2X  and  SP",  may  now  replace  the  original 
system  as  shown  in  Fig.  5 ;  and  these  may  be  combined 

y 


r 

m  FIG.  5 

into  a  single  resultant  which  is  the  diagonal  of  the  rec- 
tangle of  which  the  two  forces  are  sides  (Art.  15).  This 
gives  the  resultant  in  magnitude  and  direction,  and  this 
resultant  force  is  the  single  force  which,  if  allowed  to  act 
upon  the  material  point,  would  produce  the  same  effect  as 


12 


APPLIED  MECHANICS  FOB  ENGINEERS 


Analytically  the  resultant  may  be 


the  system  of  forces, 
expressed,  ^  = 

and  its  direction  makes  an  angle  a  with  the  re-axis  such 
that  tan«  =  ^-^.     (See  Fig.  5.)     If   the   forces   are   in 


equilibrium,  this  resultant  force  must  be  equal  to  zero; 
that  is,  V(2JT)2  +  (2F)2  =  0. 

This  means  that  (2 X)2  +  (2  F)2  =  0,  that  is,  that  the  sum 
of  two  squares  must  be  zero;  but  this  can  happen  only 
when  each  one,  separately,  is  zero  (since  the  square  of  a 
real  quantity  cannot  be  negative).  We  therefore  have  as 
the  necessary  and  sufficient  conditions  for  the  equilibrium 
of  a  material  point,  acted'  upon  by  a  system  of  concurring 
forces  in  a  plane,  ^X  =  0,  and  S  Y  =  0. 

When  R  is  not  zero,  the  system  of  forces  causes  accel- 
erated motion  in   the  direction   of  R\  when  R  =  0,  the 

material  point  remains  at 
rest,  if  at  rest,  or  continues 
in  motion  with  uniform  veloc- 
ity, if  in  motion.  In  this 
case  the  system  of  forces  is 
said  to  be  balanced. 

As  an  illustration  of  the 
foregoing,  consider  the  case 
of  a  body  of  weight  Gr  situ- 
ated on  an  inclined  plane, 
making  an  angle  6  with  the 
horizontal.  (See  Fig.  6.)  There  is  a  certain  force  P, 
making  an  angle  <£  with  the  plane,  whose  component  along 


CONCURRENT  FORCES 


13 


the  plane  acts  upwards,  and  also  a  force  of  friction  F 
upwards.  The  other  forces  acting  on  the  body  are  6r, 
the  force  of  gravity  acting  vertically,  and  N^  the  normal 
pressure  of  the  plane.  Taking  the  a>axis  along  the  plane 
positive  upward  and  the  y-axis  perpendicular  to  it  positive 
upward,  we  get, 


-  G-  sin9 

and  2F=  JV+  P  sin  </>  -  #  cos  0. 

For  equilibrium 

P  cos  <£  +  F  -  G-  sin  6  =  0, 
N+  P  sin  <£  -  a  cos  0  =  0. 
Therefore,  N=Gr  cos  0  —  P.  sin  <£, 
sin  0  -  F 


P  = 


COS0 


/ 


This  last  equation  gives  the  magnitude  of  P  required  to 
preserve  equilibrium,  supposing  that  the  force  of  friction, 
,  and  <f>  are  known. 


Problem  3.  Given  three  concurring  forces,  100  lb.,  50  lb.,  and 
200  lb.,  whose  directions  referred  to  the  z-axis  are  0°,  60°,  180°, 
respectively;  find  the  resultant  in 
magnitude  and  direction. 

v  Problem  4.  A  body  (Fig.  7)  whose 
weight  is  G  is  drawn  up  the  inclined 
plane  with  uniform  velocity  due  to 
the  action  of  the  forces  P  and  P'. 
Find  the  force  of  friction  F,  and  the 
normal  pressure  N,  if  P  =  100  lb., 
P'=100  lb.,  G=1QO  lb.  P  acts 
parallel  to  the  plane  and  P'  acts  hori- 
zontally. 


14 


APPLIED  MECHANICS  FOR   ENGINEERS 


18.  A  Body  in  Equilibrium  under  the  Action  of  Two  Forces. 
Tension,  Compression.  —  Two  forces  are  in  equilibrium  when 
and  only  when  their  lines  of  action  coincide  and  one  of  the 
forces  is  equal  in  value  but  opposite  in  direction  to  the 

other.  In  roof  and 
g —  — 3)  F>  bridge  trusses, in  cranes, 
(a)  ^zg)  (b)  and  in  other  jointed 

FlG-  8  structures  all  forces  act- 

ing on  each  member  are  often  considered  as  applied  at  two 
points  near  the  ends  of  that  member,  and  the  member  may 
thus  be  regarded  as  acted  upon  by  only  two  forces.  Where 
such  is  the  case  the  two  forces  acting  upon  the  member 
must  be  equal  and  oppositefand  their  lines  of  action  must 
coincide  with  the  line  joining  the  points  of  application  of 
the  forces. 

If  the  two  forces  act  towards  each  other,  the  member  is 
said  to  be  in  compression.  If  the  forces  act  away  from 
each  other,  the  member  is  said  to  be  in  tension.  Thus  in 
Fig.  8  the  member  is  in  compression  in  (a)  and  in  tension 
in  (6),  the  amount  of  compression  or  tension  being  P. 


FIG.  9 


Illustration.     In  Fig.  9  a  horizontal  pin  carrying  the 
weight  W passes  through  the  members  AB  and  BO  which 


CONCURRENT  FORCES 


15 


are  inclined  respectively  at  angles  60°  and  30°  to  the 
horizontal  and  held  at  A  and  Q  by  horizontal  pins  perpen- 
dicular to  the  plane  ABO.  If  the  weights  of  AB  and  BO 
are  small  compared  to  IF,  AB  and  BO  may  each  be  re- 
garded as  acted  upon  by  only  two  forces.  The  values  of 
these  forces  may  be  found  as  follows:  Since  AB  and  BO 
are  in  equilibrium  under  the  action  of  only  two  forces, 
those  acting  at  the  pins,  the  forces  acting  on  each  member 


must  be  along  the  lines  joining  the  pins.     The  members 
themselves  must  then  exert  on  the  pins  forces  equal  and 
opposite  to  those  that  the  pins  exert  on  the  members. 
Hence  acting  on  the  pin  at  B,  holding  it  in  equilibrium, 
are  the  forces  TP",  P,  and  (?,  acting  respectively  downward, 
along  AB,   and   along    OB     (Fig.   10).     Applying   the 
conditions  for  equilibrium, -2 X=  0,  2Y=  0,  there  results 
P  cos  60°  -  Q  cos  30°  =  0, 
P  sin  60°  +  Q  sin  30°  -  W=  0. 
Solving  these  equations,  we  find 

P=.866  W, 


16 


APPLIED  MECHANICS  FOE  ENGINEERS 


A  graphical  solution  is  obtained  by  laying  out  the  force 
triangle  for  the  forces  in  equilibrium  at  the  point  B  and 
measuring  the  sides  representing  P  and  Q  (Fig.  11). 

Problem  5.  A  weight  of  10  tons  is  supported  as  shown  in  Fig. 
12.  Find  the  force  acting  in  the  tie  A  and  the  member  B. 


FIG.  12 


'  Problem  6.  Two  ropes  of  length  5  ft.  and  8  ft.  are  attached  at 
one  end  to  a  weight  of  500  Ib.  The  other  ends  of  the  rope  are  at- 
tached to  two  points  9  ft.  apart  on  a  horizontal  line.  When  the 
weight  is  suspended  by  the  ropes,  find  the  tension  in  each  rope. 

19.  Body  in  Equilibrium  under  the  Action  of  Three  Forces. 
—  If  three  forces  are  in  equilibrium,  any  one  of  them  must  be 
the  equilibrant  of  the  other  two.  Its  line  of  action  must 
therefore  pass  through  the  intersection  of  the  lines  of  action 
of  the  other  two.  Also  the  vector  of  the  third  force  must 
be  equal  and  opposite  to  the  diagonal  of  the  parallelogram 
formed  on  the  vectors  of  the  other  two  forces  as  sides. 
Imposing  these  conditions  on  any  body  in  equilibrium 
under  the  action  of  three  forces  will  usually  suggest  an 
easy  method  of  finding  two  of  the  forces  acting  on  the 
body  when  the  third  one  is  known. 


CONCURRENT  FORCES 


17 


Illustration.  A  gate  9  ft.  wide  by  5  ft.  high,  weigh- 
ing 80  lb.,  is  hung  by  two  hinges  distant  respec- 
tively 6  in.  from  the  top  and  the  bottom  of  the  gate. 
If  the  lower  hinge  carries  all  of  the  weight  (i.e.  the  upper 
hinge  exerts  no  upward  force 
on  the  gate),  find  the  reactions 
of  the  hinges.  The  weight  of 
the  gate  may  be  assumed  to 
act  at  the  center. 


SOLUTION.     The  gate  is  acted  FlG<  13 

upon  by  three  forces :  the  weight,  80  lb.,  acting  down- 
ward at  the  center,  a  horizontal  force  exerted  by  the  upper 
hinge,  and  a  force  unknown  in  magnitude  and  direction 
exerted  by  the  lower  hinge.  These  three  forces  must  pass 

through  a  common  point,  the 
point  A,  Fig.  14,  and  form  a 
closed  triangle  of  forces. 
Hence,  if  AB  =  80  lb.,  the  forces 
P  and  Q  in  Fig.  14  represent  on 
the  same  scale  the  reactions  of 
the  lower  and  upper  hinges  re- 
spectively. The  values  may  be  found  by  measurement, 
or  by  solving  the  triangle  ABC.  Comparing  the  similar 
triangles  ABC  and  AED,  one  a  triangle  of  forces,  the 
other  a  triangle  of  distances,  we  have 


a  >- 
E  1— 

*  OS 

A 

' 

J/ 

80  LBS. 
r> 

\D_/^ 

<  ^-  > 

FIG.  14 


(4.5)2 


80 


.-.  P  =  120.4  lb. 


18 


APPLIED  MECHANICS  FOR    ENGINEERS 


Problems  of  this  nature  are  more  easily  solved  by  use 
of  the  theory  of  moments  developed  later. 

Problem  7.  Neglecting  the  weight  of  the  members  AD  and  BC 
in  Fig.  15,  find  analytically  and  graphically  the  stress  in  BC  and  the 
reaction  at  A . 

SUGGESTION.  BC  is  acted  upon  by  two  forces,  one  at  B  and  one 
at  C.  AD  is  acted  upon  by  three  forces.  For  the  analytical  solution 
first  find  the  values  of  AO  and  BO  and  compare  the  force  triangle 
with  the  triangle  A  OB. 


FIG.  15 


FIG.  16 


Problem  8.  A  post  20  ft.  high  is  hinged  at  the  foot  and  stands 
vertically.  Two  ropes  BC  and  ED  on  opposite  sides  of  the  post  and 
in  a  plane  perpendicular  to  the  axis  of  the  hinge  make  angles  of  50° 
and  40°  respectively  with  the  horizontal  and  are  attached  to  the  post 
at  distances  of  10  ft.  and  20  ft.  respectively  from  the  bottom.  What 
tension  in  ED  will  cause  a  tension  of  500  Ib.  in  BC,  and  what  will 
then  be  the  reaction  at  the  foot  of  the  post  ?  Solve  graphically. 

Problem  9.  Neglecting  the  weight  of  the  truss  in  Fig.  16,  find 
analytically  and  graphically  the  reaction  at  A  and  the  tension  in  BC. 

Problem  10.  Find  analytically  and  graphically  the  value  of  the 
horizontal  force  P  that  will  just  raise  the  corner  A  of  the  block  in 
Fig.  17  from  the  floor.  Find  also,  graphically,  the  value  of  P  when 
A  is  raised  6  in.  from  the  floor.  When  A  is  raised  1  ft. 


CONCURRENT  FORCES 


19 


Problem  11.  The  column  AB,  Fig.  18,  is  one  foot  square,  15  ft. 
long,  and  weighs  1200  Ib.  Find  graphically  the  tension  in  the  rope 
and  the  reaction  at  A  when  the  angle  which  the  column  makes  with 


A 


FIG.  17 

the  horizontal  is  0°,  30°,  60°.     The  pulley  at  C  is  small.    Consider 
the  weight  of  the  column  as  acting  at  its  center. 

Problem  12.  A  wheel  is  about  to  roll  over  an  obstruction.  The 
diameter  of  the  wheel  (Fig.  19)  is  3'  and  its  weight  800  Ib.  Find  the 
horizontal  force  P  through  the  center  necessary  to  start  the  wheel 
over  the  obstruction. 


FIG.  19 

Problem  13.  An  angle  iron,  whose 
weight  is  20  Ib.  and  angle  a  right  angle, 
rests  upon  a  circular  shaft,  radius  2  in. 
Find  the  normal  pressure  at  A  and  B  (Fig. 
20). 


FIG.  20 


20.   Concurrent  Forces  in  Space.  —  Let  Pv  P2,  P8,  etc.  be 
a  set  of  concurrent  forces  not  in  one  plane,  and  let  their 


20 


APPLIED  MECHANICS  FOR  ENGINEERS 


direction  angles  be  respectively  av  @v  ^ ;  c^,  /32,  72 ;  «3, 
£8,  Va ;    etc-     (The  direction  angles  of  a   force   are    the 
z 


Fio.  21 

angles  at  the  point  of  application  of  the  force  measured 
from  the  positive  direction  of  the  coordinate  axes  to  the 
vector  representing  the  force.)  The  force  P1  may  be  re- 
solved into  components  parallel  to  the  coordinate  axes, 

Xl  =  Pj  cos  «j,  Y1  =  Pl  cos  /3V  Zl  =  P1  cos  yr 
The   resultant   force   may   be   found   in   magnitude  and 
direction  by  an  analysis  similar  to  that  used  in  Art.  17. 
The  sum  of  the  components  of  all  the  forces  parallel  to 
the  ic-axis  is 

2JT=  Pl  cos  «!  4-  P2  cos  «2  +  P%  cos  «3  +  etc., 
the  sum  of  the  components  parallel  to  the  ?/-axis, 

2 T=  Pl  cos  fa  +  P2  cos  @2  +  P3  cos  /33  +  etc., 
and  the  sum  of  the  components  parallel  to  the  2-axis, 

2Z=  Pl  cos  7X  +  P2  cos  72  +  P3  cos  73  +  etc. 
The  original  system  of  forces  may  now  be  replaced  by 
a   system  of  three  rectangular  forces  2JT,  2P",  and 


CONCURRENT  FORCES  21 

(Fig.  22).  Finally,  this  system  may  be  replaced  by  a 
resultant  which  is  the  diagonal  of  a  parallelepiped  con- 
structed with  2JT,  2y,  2Z  as  edges. 
In  magnitude  this  resultant  may  be 
expressed 


(2  F  )2 

(see  Fig.  22)  and  its  direction  given  by  FlQ-  22 

the  angles  a,  &  and  y.     These  angles  are  given  by  the 

equations 

2  X  2  Y  ^7 

cos  a  =  —  —  ,  cos  /3  =  —  —  ,  cos  y  =  -- 
R  R  R 

For  equilibrium  R  must  be  0  ;  that  is, 

(2  X)2  +  (2  F)2  +  (2Z)2  =  0, 
and  therefore, 


This  gives  three  equations  of  condition  from  which  three 
unknown  quantities  may  be  determined.  In  the  preced- 
ing case  of  Art.  17  there  were  only  two  equations  of 
condition  SJT=  0  and  S  Y=  0;  consequently,  only  two  un- 
known quantities  could  be  determined. 

Problem  14.  Prove  that  if  a,  J3,  y  are  the  direction  angles  of  any 
straight  line,  then  cos2  a  +  cos2  /?  -f  cos2  y  =  1. 

Problem  15.  Three  men  (Fig.  23)  are  each  pulling  with  a  force 
P  at  the  points  a,  &,  and  c,  respectively.  What  weight  Q  can  they  raise 
with  uniform  motion  if  each  man  pulls  100  Ib.  ?  Each  force  makes 
an  angle  of  60°  with  the  horizontal  and  the  projections  of  the  forces 
on  a  horizontal  plane  make  angles  of  120°  with  each  other.  Solve 
analytically  and  also  graphically  by  projecting  each  force  on  the 
vertical  and  horizontal  planes. 


22 


APPLIED  MECHANICS  FOR  ENGINEERS 


II 

^-> 

\ 

nitude  and  c 
as  follows  : 

Ss 

y 

P1  =  75  Ib.  ;  c 

P2  =  80  Ib.  ;  c 

P3  =  95  Ib.  ;  < 

9 

1 

HINT.     yr 
the  relation  : 

Xi 

c    -TlProblem  ] 

Problem  16.     Three   concurring  forces  act 
upon  a  rigid  body.     Find  the  resultant  in  mag- 
rection.    The  forces  are  defined 

!  =  63°  27' ;  ft  =  48°  36' ;  yt  =  ? 

2  =  153°  44' ;  ft  =  67°  13' ;  y2  =  ? 

3  =  76°14';  ft  =  147°  2' ;  y3  =  ? 

y2,  and  y3  may  be  found  from 

cos2  a  +  cos2  ft  +  cos2  y  =  1. 

Each  leg  of  a  pair  of  shears 
(Fig.  24)  is  50  ft.  long.  They  are  spread  20 
ft.  at  the  foot.  The  back  stay  is  75  ft.  long. 


FIG.  23 

Find  the  forces  acting  on  each  member  when  lifting  a  load  of  20  tons 
at  a  distance  of  20  ft.  from  the  foot  of  the  shear  legs,  neglecting  the 
weight  of  the  structure. 


shear  legs, 

*  ifc  'Hi* 


FIG.  24 


21.  Moment  of  a  Force.  —  The  moment  of  a  force  with 
respect  to  any  point  is  defined  as  the  product  of  the  force 
and  a  perpendicular  from  the  point  to  the  line  of  action 


CONCURRENT  FORCES  23 

of  the  force.  Let  P  (Fig.  25)  be  the  force  and  0  the 
point  and  a  the  perpendicular  distance  of  the  force  from 
the  point ;  then  Pa  is  the  moment  of  the  force  with 
respect  to  the  point  0.  This  moment  is  measured  in 
terms  of  the  units  of  both  force  and  length,  viz.  pound- 
feet  or  pound-inches,  and  is  read  pound-feet  or  pound- 
inches  to  distinguish  it  from  foot-pounds  of  work  or 
inch -pounds  of  work. 

For  convenience  the  algebraic  sign  of  the  moment  is  said 
to  be  positive  when  the  moment  tends  to  turn  the  body  in  a 
direction    counter-clockwise,    and 
negative  when  it  tends  to  turn  the 
body  in  the  clockwise  direction. 

The  moment  may  be  repre- 
sented geometrically  as  follows : 
let  EF  represent  the  magnitude  of 
P,  drawn  to  the  desired  scale,  and 
draw  EO  and  FO.  The  area  of 
the  triangle  OEF =±  EFa,  or 

EFa  =  2  A  OEF\  that  is,  the  moment  of  the  force  with 
respect  to  a  point  is  geometrically  represented  by  twice  the 
area  of  the  triangle,  whose  base  is  the  line  representing  the 
magnitude  of  the  force  and  whose  vertex  is  the  given  point. 

If  the  moment  of  the  force  is  negative,  then  the  moment 
is  represented  by  minus  twice  the  area  of  the  triangle. 

22.  Varignon's  Theorem  of  Moments.  —  The  moment  of  the 
resultant  of  two  concurring  forces  with  respect  to  any  point 
in  their  plane  is  equal  to  the  algebraic  sum  of  the  moments 
of  the  two  forces  with  respect  to  the  same  point. 


24 


APPLIED  MECHANICS  FOR   ENGINEERS 


Let  P  and  Q  be  any  two  concurrent  forces  and  0  a 
point  in  their  plane.  Through  0  draw  a  line  parallel  to 
the  line  of  action  of  one  of  the  forces,  as  P.  Let  the  seg- 
ment A  0  cut  off  on  the  line  of  Q  by  this  line  be  taken  to 

D  C   0  T) 


A       (a) 

FIG.  26 

represent  the  force  Q,  and  let  AB  represent  the  force  P 
to  the  same  scale.  Then,  using  the  upper  sign  for  case  (a) 
and  the  lower  for  case  (5),  Fig.  26, 

mom  P  +  mom  Q  =  2  (area  of  OAB  ±  area  of  OAO). 
But  area  of  OAB  =  area  of  ABD  =  area  of  A  CD. 

.-.  2  (area  of  OAB  ±  area  of  OAO)  =  2  area  of  OAD 

=  mom  R. 
.'.  mom  P  +  mom  Q  =  mom  R. 

23.  Moment  of  a  Force  with  Respect  to  a  Line.  —  Let  P  be 
any  force  and  AB  any  line  (Fig.  27).     The  moment  of  P 

with  respect  to  AB  is  defined 
as  follows  :  Resolve  P  into 
two  components  one  of  which 
is  parallel  to,  and  the  other 
perpendicular  to,  AB.  The 
product  of  the  component  per- 
pendicular to  AB  and  the  per- 
pendicular distance  of  thiscorn- 
27  ponent  from  AB  is  called  the 


CONCUEEENT  FORCES 


25 


moment  of  P  with  respect,  to  AB.  (In  Fig.  27  the  mo- 
ment of  P  with  respect  to  AB  is  aPr  )  More  briefly  the 
definition  is :  The  moment  of  a  force  with  respect  to  a 
line  is  the  moment  of  the  projection  of  the  force  upon  a 
plane  perpendicular  to  the  line,  with  respect  to  the  inter- 
section of  the  line  and  the  plane.  The  moment  is  con- 
sidered positive  or  negative  according  as  the  tendency  to 
rotation  by  the  force  is  counter-clockwise  or  clockwise. 
The  sign  of  the  moment  will  then  change  if  the  observer 
changes  from  one  side  of  the  plane  on  which  the  projection 
is  made  to  the  other,  and  in  comparing  the  moments  of 
several  forces  with  respect  to  a  line  the  forces  should  all 
be  projected  upon  the  same  plane  and  their  projections 
viewed  from  the  same  side  of  that  plane. 

24.  Moment  of  the  Resultant  of  Two  Concurrent  Forces  with 
Respect  to  a  Line.  —  The  sum  of  the  moments  of  two  con- 
current forces  with  respect  to  any  line  is  equal  to  the  moment 
of  their  resultant  with  respect  to  that  line. 

Proof:  Let  P  and  Q  be  any  two  concurrent  forces,  R 
their  resultant,  and  AB  M 
any  line  (Fig.  28). 
Through  the  intersection 
of  P  and  Q  pass  a  plane 
MN  perpendicular  to  AB, 
cutting  ABm  0.  Project 
P,  Q,  and  R  on  MN,  the 
projections  being  respec- 
tively Pv  Qv  and  Rr  FIG.  28 
From  the  definition  of  projection  it  follows  at  once  that 
Rl  is  the  resultant  of  Pl  and  Qr  The  moments  of  P,  Q, 


26  APPLIED  MECHANICS  FOR  ENGINEERS 

and  R  with  respect  to  AB  are.  respectively  equal   to  the 
moments  of  Pv  Qv  and  Rl  with  respect  to  0.     (Definition.) 
By  Art.  22,  with  respect  to  0, 

mom  P1  +  mom  Q1  =  mom  Rr 
Therefore,  with  respect  to  AB, 

mom  P  +  mom  Q  =  mom  R. 

If  there  are  three  or  more  forces,  the  application  of  the 
above  theorem  to  the  resultant  of  two  of  the  forces  and  a 
third  force,  etc.,  proves  that  the  sum  of  the  moments  of 
any  number  of  concurrent  forces  with  respect  to  any  line  is 
equal  to  the  moment  of  their  resultant  with  respect  to  that 
line. 

COROLLARY.  From  the  definition  of  the  moment  of  a 
force  with  respect  to  a  line  it  follows  that  the  sum  of  the 
moments  of  two  forces  of  equal  numerical  value  but  oppo- 
site in  direction  and  acting  in  the  same  line  is  zero. 

Use  will  be  made  of  these  principles  in  a  later  chapter. 


CHAPTER  III 


PARALLEL  FORCES 

25.  The  Resultant  of  Two  Parallel  Forces.  —  In  consid- 
ering two  parallel  forces  three  cases  arise  :  (a)  when  the 
forces  are  in  the  same  direction  ;  (5)  when  they  are 
unequal  and  in  opposite  directions  ;  (c)  when  they  are 
equal  and  in  opposite  directions,  but  have  different  lines 
of  action. 

In  case  (<?)  the  two  forces  form  a  couple.  It  will  be 
shown  later  that  there  is  no  single  force  that  will  replace 
them. 

In  cases  (a)  and  (6)  the  two  forces  have  a  resultant. 
Its  value  and  line 
of  action  are  found 
as  follows  : 

Let  the  two 
forces  be  P1  and 
P2,  acting  at  the 
points  A1  and  A2. 
At  Al  and  A2  in 
the  line  A^A^  put 
in  two  equal  and 
opposite  forces  T^ 


T\B  T 


(a) 


FIG.  29 

and  TT  These  two  forces  will  have 
no  effect  as  far  as  the  state  of  rest  or  motion  of  the  body 
on  which  the  forces  act  is  concerned.  Combine  these 

27 


28  APPLIED  MECHANICS  FOR  ENGINEERS 

forces  with  P1  and  P2,  respectively,  obtaining  the  result- 
ants R!  and  R2.  These  resultants  may  be  moved  back 
to  the  point  of  intersection  of  their  lines  of  action  B, 
and  resolved  into  components  parallel  to  their  original 
components.  The  two  forces  T^  and  T2  then  annul  each 
other,  and  there  are  left  the  two  forces  P1  and  P2  acting 
at  B  parallel  to  their  original  positions.  The  resultant  then 
in  case  (#)  is  R  =  Pl  +  P2,  and  in  case  (5)  R  =  P2  —  Pr 

Let  the  line  of  action  of  the  resultant  cut  the  line  A±AZ 
in  O,  and  let  the  distances  from  0  to  A-±  and  A^  be  re- 
spectively d1  and  d2. 

Then  from  similar  triangles,  in  either  case, 

BC     P        ,  BO     Po 


By  division,  remembering  that  T^  =  Ty 


Hence,  the  resultant  of  two  parallel  forces  acting  in  the 
same  direction  is  equal  to  their  sum,  is  parallel  to  the  forces, 
and  divides  the  line  joining  their  points  of  application  in 
the  inverse  ratio  of  the  forces. 

The  resultant  of  two  unequal  parallel  forces  acting  in 
opposite  directions  is  equal  to  their  difference,  acts  in  the 
direction  of  the  larger  force,  and  divides  the  line  joining 
their  points  of  application  externally  in  the  inverse  ratio  of 
the  forces. 

26.  The  Moment  of  the  Resultant  of  Two  Parallel  Forces. 
—  Applying  the  theorem  of  Art.  24  for  the  moment  of 


PARALLEL  FORCES  29 

the  resultant  of  concurrent  forces  to  the  forces  of  the 
preceding  article, 

mom  R  =  mom  Rl  +  mom  R^ 

=  mom  P1  +  mom  T^  +  mom  P2  +  mom  T2. 
But  mom  T±  4-  mom  Tz  =  0  (Art.  24,  Cor.). 

Therefore  mom  R  =  mom  P^  +  mom  .P2. 

Or,  the  moment  of  the  resultant  of  two  parallel  forces  with 
respect  to  any  line  is  equal  to  the  algebraic  sum  of  the  mo- 
ments of  the  two  forces  with  respect  to  that  line. 

27.  The  Resultant  of  Any  Number  of  Parallel  Forces  in 
Space.  —  By  combining  the  resultant  of  two  parallel  forces 
with  a  third,  and  that  resultant  with  another,  and  so  on, 
the  theorems  of  the  two  preceding  articles  may  be  ex- 
tended at  once  to  any  number  of  parallel  forces : 

(1)  Except  when  it  is  a  couple,  the  resultant  of  any  number 
of  parallel  forces  in  space  is  equal  to  their  algebraic  sum  and 
acts  parallel  to  the  forces. 

(2)  The  algebraic  sum  of  the  moments  of  any  number  of 
parallel  forces  in  space  with  respect  to  any  line  is  equal  to 
the  moment  of  their  resultant  with  respect  to  that  line. 

It  follows  at  once  that  if  a  set  of  parallel  forces  is  in 
equilibrium,  the  sum  of  the  moments  of  all  the  forces  with 
respect  to  any  line  is  zero,  since  the  resultant  of  the  forces 
is  zero,  and  hence  the  moment  of  the  resultant  is  zero. 

If,  conversely,  the  sum  of  the  moments  of  a  set  of  par- 
allel forces  with  respect  to  a  line  not  parallel  to  the  forces 
is  zero,  either  the  resultant  of  the  forces  must  be  zero, 
or  else  the  line  of  action  of  the  resultant  must  intersect 
the  given  line.  If,  however,  the  sum  of  the  moments 


30 


APPLIED  MECHANICS  FOR   ENGINEERS 


of  the  forces  about  each  of  two  parallel  lines  in  a  plane 
not  parallel  to  the  lines  of  action  of  the  forces  is  zero, 
the  resultant  must  be  zero  and  the  forces  are  in  equi- 
librium. 

If  the  lines  of  action  of  the  forces  all  lie  in  one  plane, 
the  most  convenient  line  about  which  to  take  moments  is 
perpendicular  to  the  plane  of  the  forces. 

The  condition  for  equilibrium  in  this  case  may  be 
stated  :  If  the  sum  of  the  moments  of  a  set  of  parallel 
forces  in  one  plane  with  respect  to  each  of  three  points  of 


FIG.  30 

the  plane  not  in  the  same  straight  line  is  zero,  the  forces  are 
in  equilibrium. 

Illustration.  Forces  of  7  lb.,  5  lb.,  and  10  Ib.  act  per- 
pendicular to  the  plane  of  the  lines  OA  and  OB,  as  shown 
in  Fig.  30.  The  distances  of  the  lines  of  action  of  the 
forces  from  OA  and  OB  are  respectively  1,  3,  4  units  and 
2,  1,  5  units.  To  find  where  the  line  of  action  of  the 
resultant  cuts  the  plane  OAB : 

Let  a  and  b  be  the  distances  of  the  line  of  action  of  the 
resultant  from  OA  and  OB  respectively.  The  result- 

antis  .#=7+5  +  10  =  221b., 

acting  in  the  direction  of  the  forces. 


PARALLEL   FORCES 


31 


Taking  moments  about  the  line  OA, 

a  .  E  =  1  •  7  +  3  •  5  +  4  . 10, 
or  a  =  2.82  units  of  space. 

Taking  moments  about  OB, 

b.  J2  =  2. 7  +  1-5+5.  10, 
or  b=  3.14  units  of  space. 

Hence  the  resultant  is  equal  to  22  lb.,  acts  in  the  direc- 
tion of  the  forces,  and  at  distances  of  2.82  and  3.14 
units  from  OA  and  OB  respectively. 


\ 


FIG.  31 


Problem  18.  Two  parallel  forces,  one  of  20  lb.  and  one  of  100  lb., 
have  lines  of  action  24  in.  apart.  Find  the  resultant  in  magnitude, 
direction,  and  line  of  action  : 

(1)  When  they  are  in  the  same  direction. 

(2)  When  they  are  in  opposite  directions. 

Problem  19.  A  horizontal  beam  of  length  I  is  supported  at  its 
ends  by  two  piers  and  loaded  with  a  single  load  P  at  a  distance  of 

-  from  one  end.     Find  the  reactions  of  the  piers  against  the  beam. 

Problem  20.  The  locomotive  shown  in  Fig.  31  is  run  upon  a 
turntable  whose  length  is  100  ft.  Find  the  position  of  the  engine 
so  that  the  table  will  balance. 


32 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  21.     Weights  of  100,  50,  and  120  Ib.  are  placed  at  the  ver- 
tices A,  B,  and  C  respectively  of  an  equilateral  triangle  4  ft.  on  a  side. 
Find  the  distances  of  the  center  of 
the  forces  from  the  lines  AB  and  BC. 

Problem  22.  On  a  square  table 
weights  of  70,  80,  and  100  Ib.  are 
placed  as  shown  in  Fig.  32.  Find  the 
position  of  a  fourth  weight  of  90  Ib. 
which  will  balance  the  given  weights 
with  respect  to  the  two  lines  AB  and 
CD. 

Ans.     f  ft.  from  AB,  3$  ft.  from  CD  in  quadrant  BOD. 


FIG.  32 


28.  Center  of  Parallel  Forces.  —  In  Art.  25  it  was  shown 
that  the  resultant  of  two  parallel  forces  P1  and  P2,  acting 
at  the  points  A1  and  A2  respectively,  divides  the  line  A-^A2 
in  the  inverse  ratio  of  the  forces.  If,  then,  without 
changing  the  points  of  application  of  the  parallel  forces 


/L          / 

Jr     i 

i 

i 
i 

i 
i 
i 

i 

i 

i 
i 

t 

cj 

i 
i 

R 

Pl+p,   " 


FIG.  33  FIG.  34 

or  their  magnitudes  their  direction  be  changed,  the  point 
on  A1A^  through  which  the  resultant  passes  is  not  changed, 
since  it  must  still  divide  A1A2  in  the  ratio  P2 :  Pr 

In  the  same  way,  if  three  parallel  forces  of  fixed  value 
Pj,  P2,  P3  act  at  three  fixed  points  Av  A2,  A3,  and  if  O' 


PARALLEL   FORCES 


33 


is  the  point  on  A^A^  through  which  the  resultant  of  P1 
and  P2  passes,  then  the  point  C  on  O'A3  which  divides 
CfAQ  in  the  ratio  PB :  (Pj  +  P2)  is  a  point  through  which 
the  resultant  of  the  three  forces  always  passes  no  matter 
what  their  direction.  The  same  reasoning  applies  to  any 
number  of  parallel  forces  acting  at  definite  points.  For 
any  set  of  parallel  forces  acting  at  definite  points  there  is 
therefore  a  point  through  which  the  resultant  always 
passes  no  matter  what  direction  the  parallel  forces  may 
take.  This  point  is  called  the  center  of  the  parallel  forces. 
29.  Graphical  Construction  for  the  Center  of  Parallel 
Forces.  —  Let  the  parallel  forces  Pl  and  P2  act  at  Al  and 


Az  (Fig.   35).     From  Al  lay  off 


equal  to  P2  but 


lay  off  A2B2 


in  the  opposite  direction  to  P2.  From 
equal  to  Pl  and  in  the  same  direction  as  Pr  The  inter- 
section of  B^BI  and  A^2  is  then  the  center  of  the  two 
forces.  For,  letting  A1O  =  d1  and  CA2  =  dv  we  have,  by 
similar  triangles  A^Bfi  and  A2B2C, 

4  =  dA,    or     i  =  :*J. 
^2     A2B2  d2     Pl 

Hence  C  is  the  point  found  in  Art.  25  to  be  the  center  of 
the  forces. 


34  APPLIED  MECHANICS   FOR  ENGINEERS 

If  there  are  more  than  two  parallel  forces,  the  continued 
application  of  the  construction  will  locate  the  center  of 
the  forces. 

Problem  23.  Weights  of  20,  45,  and  70  Ib.  are  suspended  from 
points  on  a  straight  line  at  distances  of  2,  5,  and  7  ft.  respectively 
from  a  point  O.  Find  by  graphical  construction  the  center  of  the 
forces  exerted  by  the  weights.  Check  by  the  theory  of  moments. 

Problem  24.  Locate  graphically  the  center  of  the  three  forces  in 
the  illustration  of  Art.  27.  (N.B.  The  forces  may  be  assumed  to 
act  in  the  plane  OAB  without  changing  the  position  of  the  center.) 

30.  In  Eegard  to  Signs.  —  Given  a  set  of  parallel  forces, 
Pv  P2,  P3,  etc.,  acting  at  definite  points. 

Let  OT  be  a  line  perpendicular  to  the  lines  of  action 
of  the  forces.  Let  it  be  agreed  that  forces  acting  in  one 
direction,  as  Pv  Fig.  36,  shall  be  positive  and  those  acting 
in  the  opposite  direction  negative. 

Let  xv  z2,  #3,  ...  be  the  perpendiculars  from  OY  to  the 

lines  of  the  forces Pv  P2,  P3,  •.., 
x  being  counted  positive  if 
measured  from  the  axis  OY  to- 
ward the  line  of  a  positive  force 
which  tends  to  cause  counter- 
clockwise rotation  to  an  ob- 
server looking1  from  Y  to  0  in 
FIG.  36 

Fig.  36,  and  negative  in  the  op- 
posite direction.  Then,  when  observed  in  the  direction 
YO,  Px  is  the  moment  of  P  with  respect  to  OY,  not  only 
in  numerical  value  but  in  sign.  For  if  both  P  and  x  are 
positive,  as  Pl  and  xv  or  both  negative,  as  P3  and  #3,  the 
product  is  positive.  The  tendency  to  rotation  in  both 
cases  is  counter-clockwise,  or  positive, 


PARALLEL  FORCES 


35 


If  x  is  positive  and  P  is  negative,  as  xz  and  P2,  or  if  x 
is  negative  and  P  is  positive,  as  #4  and  P4,  the  product  is 
negative,  and  an  inspection  of  the  figure  shows  that  the 
tendency  to  rotation  is  clockwise,  which  has  been  defined 
as  negative. 

The  sum  of  the  moments  of  the  forces  with  respect  to 
OF  is  then  P^  +  P2x2  +  P3z3  +  P4#4  +  •••.  This  sum 
is  denoted  by 


31.  Coordinates   of  the  Center   of   a    System    of    Parallel 
Forces. — Let   Pr   P2,  P3,  ...  be  a  set  of   parallel  forces 
whose  points  of  applica- 
tion are  (a?1,  y^  z-^),  (x^ 

rectangular  coordinates. 
Denote  the  center  of  the 
forces  by  (#,  y,  z).  As 
was  shown  in  Art.  28  the 
position  of  the  center  of 
the  forces  is  independent 
of  the  direction  in  which 
they  act.  Assume  then 
that  they  act  parallel  to  OZ. 

Taking  moments  about  OY,  we  have  the  moment  of 
the  resultant  equal  to  the  sum  of  the  moments  of  the 
forces,  or 


A. 


FIG.  37 


or 

Taking  moments  about  OX, 


86 


APPLIED  MECHANICS  FOR  ENGINEERS 


Assuming  the  forces  to  act  parallel  to  OX  and  taking 
moments  about  OY,  we  obtain 


Therefore 


Problem  25.  Forces  of  40,  65,  and  70  Ib.  act  in  the  same  direction 
from  the  points  (4,  3,  1),  (5,  -2,  3),  and  (2,  3,  6),  respectively. 
Find  the  resultant  and  the  center  of  the  forces. 

Problem  26.  Three  equal  parallel  forces  act  at  the  vertices  of 
a  triangle  in  the  same  direction;  prove  that  their  resultant  acts 
at  the  intersection  of  the  medians  of  the  triangle.  Solve  by  applying 
the  theorem  of  moments,  and  check  by  graphical  construction. 

Problem  27.  Four  equal  forces  act  in  the  same  direction  at  the 
vertices  of  a  regular  tetrahedron.  Find  the  center  of  the  forces  by 
taking  moments.  Also  locate  the  center  graphically. 

32.  Arrangement  of  the  Work  for  Computation.  —  If  sev- 
eral parallel  forces  and  their  points  of  application  are 
given,  it  is  sometimes  worth  while  to  tabulate  the  work 
in  some  form  such  as  the  following : 


FORCES 


Resultant 


COORDINATES 


ar, 


y* 


MOMENTS 


Px 


Py 


Pz 


PARALLEL   FORCES 


37 


P3,  and    P4  act    at   the 


The  values  of  quantities  in  the  columns  one,  two,  three, 
and  four  are  given  except  in  the  last  row.  The  values  of 
the  quantities  in  columns  five,  six,  and  seven  are  then 
computed  and  their  sums  and  the  sum  of  column  one 
placed  at  the  foot.  The  values  of  #,  y,  and  z  are  then 
easily  computed. 

Problem  28.  Using  the  above  arrangement,  find  the  coordinates 
of  the  center  of  the  following  parallel  forces :  Pl  =  50  lb.,  P2  =  100  lb., 
P3  =  300  lb.,  P4  =  10  lb.,  and  P5  =  -  400  lb.,  the  points  of  applica- 
tion being  respectively  (2,  1,  -5),  (-1,  -2,  4),  (2,  1,  -2),  (-2,  1, 1), 
(1,1,1).  Ans.  (3,  -4,  -14). 

Problem  29.  Parallel  forces  Pv  P 
corners  of  a  rectangle  3  ft.  by  2  ft. 
and  perpendicular  to  its  plane.  Find 
the  point  of  application  of  the  result- 
ant, if  PJ  =  10  lb.,  P2  =  50  lb.,  P3  = 
100  lb.,  P4  =  200  lb.,  Pjand  P2  be- 
ing 2  ft.  apart,  and  P3  on  same  side 
as  P2. 

Problem  30.    Eight  parallel  forces 
act  at  the  corners  of  a  one-inch  cube. 
Find  the  point  of  application  of  the    y/ 
resultant  force,  if  Pl  =  30  lb.,  P2  =  50 
lb.,  P3  =  10  lb.,  P4  =  20  lb.,  P5  =  100 

lb.,  P6  =  5  lb.,  P7  =  10  lb.,  P8  =  40  lb.  The  subscripts  of  the  forces 
acting  at  the  various  vertices  are  shown  in  the  figure  (Fig.  38). 

Ana.  (.283,  .302,  .415). 


FIG.  38 


CHAPTER   IV 
CENTER  OF  GRAVITY 

33.  Definition  of  the  Center  of  Gravity.  —  The  center  of 
gravity  of  a  body  may  be  defined  as  the  point  of  appli- 
cation of  the  resultant  attraction  of  the  earth  for  that 
body,  and  the  center  of  gravity  of  several  bodies  consid- 
ered together,  as  the  point  of  application  of  the  resultant 
attraction  of  the  earth  for  the  bodies.  The  attention  of 
the  student  is  called  to  the  fact  that  the  forces  acting 
upon  the  particles  of  a  body,  due  to  the  attraction  of  the 
earth,  are  not  parallel,  but  meet  in  the  center  of  the  earth. 
For  all  practical  purposes,  however,  they  are  considered 
parallel. 

The  center  of  gravity  of  many  simple  bodies  can  be 
found  by  inspection.  For  example,  for  a  sphere  of  uni- 
form material  it  is  evident  that  the  line  of  action  of  the 
earth's  attraction  always  passes  through  the  center.  For 
a  rectangular  block  of  uniform  material  the  same  is  true. 

If  a  body  be  divided  up  into  a  number  of  parts  whose 
centers  of  gravity  are  known,  the  whole  weight  of  the 
body  is  the  resultant  of  the  known  weights  of  the  parts 
acting  at  known  points,  and  hence  the  theorem  of  moments 
may  be  applied  to  determine  the  position  of  the  center  of 
gravity  of  the  body  ;  i.e.  the  equations 


CENTER   OF  GRAVITY 


39 


may  be  used  where  Pv  P2,  etc.,  represent  the  weights  of 
the  known  parts,  xv  #2,  etc.,  the  abscissas  of  the  centers  of 
gravity  of  these  parts,  etc. 

As  an  illustration  consider  the  prob- 
lem of  finding  the  center  of  gravity  of 
the  solid  shown  in  Fig.  39. 

The  figure  represents  a  Z-iron  of  the 
same  cross  section  throughout,  and  Pv 
P2,  and  P3  are  the  weights  of  the  indi- 
vidual parts  (considering  the  Z-iron  as 
divided  into  three  parts  —  two  legs  and 
the  connecting  vertical  portion).  If 
the  weight  of  a  cubic  inch  of  iron  =  .26 
lb.,P1  =  .78  lb.,P2=2.08  lb.,P3  =  1.04 
lb.,  and  therefore  .#  =  3.9  lb.  The 
points  of  application  of  Pv  P2,  and  P3  are  (—  |,  —  -J,  9J), 
(i>  ~  lb  ^)i  anc^  (^'  —  i"'  i)>  respectively,  so  that 

1.04(2)  _ 


-0+2.08(1) 
3.9 


= 


3.9 

2.08(5)+  1.04(1) 


-  .50  in., 


3.9 


This  point  #,  ?/,  z  is,  in  this  case,  the  center  of  gravity 
of  the  Z-iron. 

34.  Centers  of  Gravity  of  Uniform  Bodies  and  of  Areas.  —  If 
Px,  P2,  P3,  etc.,  are  the  weights  of  the  parts  of  a  body  or 
of  several  bodies,  whose  unit  weights  are,  respectively, 
7r>  ^2'  73'  e^c-'  an(^  volumes  Vv  V^  F"3,  etc.,  we  may 


40  APPLIED  MECHANICS  FOR   ENGINEERS 


write  yjFi  for  Pp  72F2  for  P2,  y8F8  for  P3,  etc.     The 
formulae  for  a;,  ?/,  2  then  become 


~  -  7i        i  +  72 


And  if  the  bodies  are  all  of  the  same  material  and  so  have 
the  same  heaviness,  y  is  constant  and  may  be  taken  outside 
the  summation  sign,  where  it  cancels  out.  This  gives 
values  for  #,  y,  and  2, 

-     SFa>     -     SF?/     -     ZFz 


35  —     <*•  *r    '       2/   —     v-^r'      S  ~     ^F   ' 


formulae  exactly  similar  to  those  of  Art.  33,  where  the  P's 
are  replaced  by  F's. 

If  the  bodies  are  thin  plates  of  the  same  material,  of 
constant  thickness  5,  we  may  write  for  F^  F^,  Fg,  etc. 
bFv  IFV  IFfr  etc.,  where  the  P's  represent  the  areas  of 
the  faces  of  the  plates.     Making  this  substitution  for  the 
F's,  x,  y,  z  may  be  written 

etc. 


etc. 


= 


= 

~  2P  ' 

the  5,  being  a  constant  factor,  cancels  out.  These  formulae 
are  taken  as  defining  the  "  center  of  gravity  of  an  area  " 
and  are  much  used  by  engineers  for  finding  the  center  of 
gravity  of  sections  of  angles,  channels,  T-sections,  Z-sec- 
tions,  etc.  Usually,  the  #?/-plane  is  taken  in  the  plane  of 
the  area  and  only  the  values  of  x  and  y  are  needed. 


CENTER   OF  GRAVITY 


41 


•  0.4' 


Problem    31. 

Find  the  center 
of  gravity  of  the 
channel  section 
shown  in  Fig.  40. 

Problem    32. 

Find   the   center 

of  gravity  of  the   T-section 

shown  in  Fig.  41. 


FIG.  40 


FIG.  41 


FIG.  42 


Problem  33.  Find  the  center  of  gravity]  of  the  U  -section  shown 
in  Fig.  42.  Given  the  fact  that  the  center  of  gravity  of  a  semicircular 

area  is  —  —  from  the  diameter.     (See  Prob.  41.) 
3  TT 

Problem  34.  Find  the  position  of  the  center  of  gravity  .of  a 
trapezoidal  area,  the  lengths  of  whose  parallel  sides  are  ^  and  a2,  re- 
„  „  spectively,  and  the  distance 


FIG.  43 


between  them  h.     (See  Fig. 
43.) 

HINT.  Draw  the  diago- 
nal AB  and  call  the  tri- 
angle ACS,  Fv  and  the 
triangle  ABD,  F2.  Given, 
the  distance  from  the  base  to 


the  center  of  gravity  of  a  triangle  is 

the  vertex.     (See  Prob.  39.)     Select  A  D  as  the  x-axis,  then 


42 


APPLIED  MECHANICS  FOR   ENGINEERS 


_ 

FI      * 

where  yl  =  f  h  and  yz  =  %  h.     The  center  of  gravity  is  seen  to  lie  on  a 

line  joining  the  middle  points  of  the  parallel  sides. 

Problem  35.     A  cylindrical,  piece  of  cast  iron,  whose  height  is 

6  in.  and  the  radius  of  whose  base  is  2  in.,  has  a  cylindrical  hole  of  1  in. 
radius  drilled  in  one  end,  the  axis  of  which 
coincides  with  the  axis  of  the  cylinder.  The 
hole  was  originally  3  in.  deep,  but  has  been 
filled  with  lead  until  it  is  only  1  in.  deep. 
Find  the  center  of  gravity  of  the  body,  the  unit 
weight  of  lead  being  710  and  of  cast  iron  450 
(Fig.  44). 

Problem  36.     Find  the  center  of  gravity  of 
a  portion  of  a  reinforced  concrete  beam.     (See 
Fig.  45.)     The  beam  is  reinforced  with  three 
half -inch  steel  rods,  centers  1  in.  from  the  bottom  of  the  beam  and  1 


4rl 


J__ 


FIG.  44 


d- 


FIG.  45 


CENTER   OF  GRAVITY 


43 


in.  from  the  sides.     The  center  of  the  middle  rod  is  4  in.  from  the 
sides. 

(y  for  steel  =  490  Ib.  per  cubic  foot ; 
y  for  concrete  =  125  Ib.  per  cubic  foot.) 

NOTE.     It  is  seen  that  the  thickness  cancels  out  of  the  expression 
for  the  center  of  gravity,  and  might,  therefore,  have  been  neglected. 

35.   Center  of  Gravity  of  a  Body  with  a  Portion  Removed.  — 

It  is  sometimes  convenient  in  finding  the  center  of  gravity 
of  a  body  to  regard  it  as  a  larger  body  from  which  a  por- 
tion has  been  removed.  The  weight  of  the  given  body 
can  then  be  regarded  as  the  resultant  of  the  weight  of  the 
larger  body  acting  at  its  center  of  gravity  and  a  force 
equal  to  the  weight  of  the  portion 
removed  acting  at  its  center  of 
gravity  in  the  opposite  direction. 

To  illustrate  this  consider  the 
cylinder  of  Fig.  46  of  height  H  and 
radius  R  from  which  a  cylindrical 
portion  of  height  h  and  radius  r  and 
having  the  same  axis  is  removed 
from  one  end. 

If  W  is  the  weight  of  the  whole 
cylinder,  and  w  the  weight  of  the  portion  removed,  then 
the  center  of  gravity  of  the  remaining  body  is  the  center 
of  the  two  forces  W  and  w  acting  as  shown  in  the  figure. 
If  7  is  the  heaviness, 

W=  yjrWJI, 
w  =  j7rr2h. 

The   distance   y   from    the   open    end   to   the   center   of 
gravity  of  the  body  is  therefore 


FIG.  46 


44        .      APPLIED  MECHANICS  FOR  ENGINEERS 

ry7rlt? 
_  = 


~  2 
In  general  the  formula  is, 


where  TF^  is  the  weight  removed  from  the  weight  Wl 
and  x1  and  x2  are  the  abscissas  of  the  centers  of  gravity 
of  the  weights  W1  and  TT2,  respectively,  before  the  removal 
of  the  weight  TF2. 

Problem  37.  Through  a  circular  disk  1  ft.  in  diameter  a  hole 
4"  in  diameter  is  bored  with  its  axis  distant  3"  from  the  axis  of  the 
disk.  Find  the  position  of  the  center  of  gravity  of  the  remainder. 

Problem  38.  From  two  adjacent  corners  of  a  cube  of  edge  1  ft. 
cubes  of  2-inch  and  3-inch  edges  are  removed.  Find  the  center  of 
gravity  of  the  remaining  portion. 

36.  Center   of   Gravity   Determined   by   Integration.  —  In 

many  cases  of  areas  and  solids  the  position  of  the  center 
of  gravity-  may  be  determined  by  integration.  The  method 
as  applied  to  areas  is  as  follows : 

Let  F  be  any  area  in  the  ^-plane.  In  order  to  deal 
with  actual  forces  think  of  F  as  the  face  of  a  thin  sheet 
of  uniform  thickness  and  uniform  material  and  let  7  be 
the  weight  of  the  sheet  per  unit  area  in  the  surface  F. 
Divide  the  area  up  into  elements  of  area  AJ7.  (This  may 
be  done  in  a  variety  of  ways,  Fig.  47.)  The  weight  of 
such  an  element  is  then  <y&F.  The  weight  of  the  sheet 
may  then  be  replaced  by  a  set  of  parallel  forces  of  magni- 
tude yAF  acting  at  the  centers  of  these  elementary  areas. 


CENTER   OF  GRAVITY 


45 


The  centers  of  the  elementary  areas  and  the  elementary 
areas  themselves  may  be  determined  approximately,  and 
if  xf  is  the  abscissa  of  the  approximate  center  *  of 


(c) 


then  an  approximate  value  for  the  abscissa  of  the  center 
of  the  parallel  forces  is 

' 


The  limiting  value  of  xr  as  the  elementary  areas  are  in- 
definitely decreased  is  defined  as  the  abscissa,  £,  of  the 
center  of  gravity  of  the  area  F.  Hence, 


Lim 


as  A.F  approaches  the  limit  zero.     By  a  theorem  of  calculus 
this  may  be  written 

Cx'ydF 

r      _  •-/ 


*The  approximate  center  must  of  course  be  such  that  the  approximate 
center  and  the  true  center  of  gravity  of  the  element  have  the  same 
limiting  position  as  the  elementary  area  is  indefinitely  decreased.  If 
both  dimensions  of  AF  approach  the  limit  zero  as  in  (6)  and  (c),  Fig.  47, 
the  approximate  center  may  be  taken  as  any  point  of  AF. 


46 


APPLIED    MECHANICS  FOR   ENGINEERS 


or,  since  7  is  constant, 


x'dF 


f 


dF 

the  integration  being  taken  to  include  the  whole  area  F. 
Similarly 

'y'dF 


£ 

y  =  ^— 

/• 


dF 

In  deriving  the  latter  formula  the  forces  would  be 
thought  of  as  acting  parallel  to  the  #-axis. 

In  the  use  of  these  formulae  it  must  be  kept  in  mind 
that  x'  and  y1  are  the  coordinates  of  the  approximate 
center  of  the  elementary  area  AF. 

Illustration.     To  find  the  center  of  gravity  of  the  area 

bounded  by  the 
o>axis,  the  curve 
y  =  sin  #,  and  the 

line  z=Z* 


i 


FIG.  48 


Divide  the  area 
up  into  strips 
L  parallel  to  the 
?/-axis  of  width 
Az.  Then  con- 
sidering any  strip  whose  middle  ordinate  intersects  the 
curve  in  the  point  (x,  y),  the  value  of  AJ7  is  z/Az  ap- 
proximately, the  value  of  x'  is  x,  and  the  value  of  y1  is 

:•  48). 


CENTER   OF  GRAVITY 


47 


Then 


_  z 

rxydx       I    x  sin  xdx 
_ Jo 

—  —  — .  — 

7T 

Jydx         C2sin  xdx 
Jo 


(—  XGOSX+  sin  a;)  |       1 
= k  =  ± 

— cosx  r 

Xy~  ydx      -  I     sin2  xdx 
—  2  2*^0 

X2  S*2    . 

ydx          I    sin  xdx 
Jo 


—  cosx 


dx-\ 


1 


2        7T 

0_    8_7T 

"~1~8 


FIG.  49 

Problem  39.  Find  the  center  of  gravity  of  a  triangle  whose  alti- 
tude is  h  and  whose  base  is  a.  Take  the  origin  at  the  vertex  and 
draw  the  x-axis  perpendicular  to  the  base.  (See  Fig.  49.) 


48 


APPLIED  MECHANICS  FOR  ENGINEERS 


x  =  -J— Here  dF  =  a'dx,  and  from  similar  triangles 

(dF 


and 


*\' 
2J, 


The  center  of  gravity  is  |  the  distance  from  the  vertex  to  the  base, 
and  since  the  median  is  a  line  of  symmetry,  it  is  a  point  on  the 
median.  It  is,  in  fact,  the  point  where  the  medians  of  the  triangle 
intersect. 


FIG.  60 

Problem  40.     Find  the   center   of  gravity  of   a  parabolic  area 
shown  in  Fig.  50,  the  equation  of  the  parabola  being  y2  =  2  px. 
Here  dF  =  ydx,  so  that 


_a 


It  is  left  as  a  problem  for  the  student  to  show  that  y  =  f  h. 

Problem  41.   Find  the  center  of  gravity  of  a  sector  of  a  flat  ring 
outside  radius  Rl  and   inside   radius  R2.     (See  Fig.  51.)     Let  the 


CENTER   OF  GRAVITY 


49 


angle  of  the  sector  be  2  0.     Take  the  origin  at  the  center  and  let  the 
x-axis  bisect  the  angle  2  0. 

Here  dF  =  pdpda,  and  x  —  p  cos  a,  so  that 

f  f  cos  ada  •  p2dp 

^  pdpda 


FIG.  51 

Integrating  the  numerator  first, 

cos  ada  — 


(2  sin  0). 


Integrating  the  denominator, 


Therefore, 


6 


If  Rz  =  0,  the  sector  becomes  the  sector  of  a  circle,  and  x  becomes 

_2       sin  0 
-"" 


If  the  sector  is  a  semicircle,  that  is,  if  2  0  =  TT,  then,  since  0  = 


50 


APPLIED  MECHANICS  FOR   ENGINEERS 


Problem  42.   Find  the  center  of  gravity  of  a  semi-ellipse  (Fig.  52) 
whose  equation  is 

' 


dF  = 


=  2-a*-x*.  dx, 
a 


therefore 


-x*dx 


aJo 


a2   TT      3  IT 

2  '2 


CENTER   OF  GRAVITY 


51 


V   Problem  43.     Find  the  center  of  gravity  of  the  area  between  the 

parabola,  they-axis,  and  the  line  AB  in  Problem  

40. 


Problem  44.   A  quadrant  of  a  circle  is  taken 
from  a  square  whose  sides  equal  the  radius  of  the  cir- 
cle.   (See  Fig.  53.)    Find  the  center 
of  gravity  of  the  remaining  area. 

Problem  45.    Suppose  that  the 
corners  A  and  C  of  the  angle  iron 


Y 


*—  i- 


\ 


FIG.  53 


in  Fig.  54  are  cut  to  the  arc  of  a  circle  of  •£$  in.  radius 
and  the  angle  at  B  is  filled  to  the  arc  of  a  circle  f  in. 
radius;  what  would  be  the  change  in  ~x  an 


37.  Center  of  Gravity  of  a  Solid. 
—  By  dividing  a  solid  into  in- 
finitesimal parts  and  taking  the 
moments  of  the  weights  of  these 
parts  with  respect  to  the  three 

coordinate  axes,  the  coordinates  of  the  center  of  gravity 

of  the  solid  may  be  found. 

As  an  illustration,  suppose  it  is  desired  to  obtain  the 

center  of  gravity  of  a  right  circular  cone  of  altitude  h  and 

radius    of    base   r. 

Take  the  #-axis  as  z 

the  axis  of  the  cone 

with  the  vertex  at 

the  origin.       (See 

Fig.    55.)       It   is 

evident  that  y  =  0 

and  2  =  0,  so  that 

it  is  only  necessary 

to    find    x.      The  FIG.  55 


52  APPLIED  MECHANICS  FOR  ENGINEERS 

volume,  dv,  cut  from  the  cone  by  two  parallel  planes,  per- 
pendicular to  OJTand  separated  by  a  distance  dx,  is  7r 
and  the  weight  of  this  dv  is  ^iry^dx  —  dP.     Therefore 


/WP    f 
x=J—^ mj± 


But  from  similar  triangles  y  :  x : :  r  :  h  or  y  =  -x.     This 
gives 

yrr—  I   z?dx 
/i  *sQ  4 


^T 

a  1 


The  expressions  for  a?,  y,  and  z,  involving  c?P,  may  be 
changed  to  similar  ones  involving  dv,  and  these  become 
for  homogeneous  bodies,  since  dP  =  ydv, 


f 


z'dv 


~ 

J 


cfo 


The  center  of  gravity  of  thin  homogeneous  wires  of  con- 
stant cross  section  may  be  found  by  replacing  the  dv  in 
the  above  formulae  by  ads,  where  a  is  the  constant  area 
of  cross  section  and  ds  is  a  distance  along  the  curve.  The 
formulae  then  become 

Cx'ds  Cy'ds  Cz'ds 

x=^-~—,   y  =  J-c—^    *=*L~  — 

fds  fds  fds 

Problem  46.  Find  the  center  of  gravity  of  a  hemisphere,  the 
radius  of  the  sphere  being  r.  Let  the  equation  of  the  generating 
circle  of  the  surface  be  x2  +  z/2  =  r2.  Divide  the  hemisphere  into 


CENTER   OF  GRAVITY 


53 


slices  of  thickness  dx  by  planes  parallel  to  the  plane  face  of  the 
hemisphere. 

(x'dP 
=  *-  - 


Then 


Show  that 


,  where  dP  =  yiry*dx  =  yir(r*  -  x^dx. 


FIG.  56 


Problem  47.  Find  the  center  of  gravity  of  a  portion  of  circular 
wire  (Fig.  56)  of  length  L  and  whose  chord  =  2  b.  Take  the  center 
of  the  circular  arc  as  origin  and  let  the  a>axis  bisect  L.  Then 


f 


xds 


$ 


ds 


But  x2  +  2/2  =  R2 

.-.  2  xdx  +  2  ydy  =  0. 


1-dy 


x  — 


arc 


54 


APPLIED  MECHANICS  FOR   ENGINEERS 


For  a  semicircular  wire 


-  _  Diameter 


Problem  48.     Find  the  center  of  gravity  of  a  paraboloid  of  revo- 
ution.     If  the  equation  of  the  generating  curve  is  y2  =  2px,  and  the 
greatest  value  of  x  is  a,  show  that 


SUGGESTION.     Use  the  same  method  as  that  used  for  the  right 
circular  cone. 

Problem  49.     (a)  Show  that  the  center  of  gravity  of  the  circular 
sector  A  OB  (Fig.  57)  of  angle  2  a  and  chord  2  d  is  given  by 

—  _  2  d  _  2  radius  x  chord 
X~3a~3~        arc 


FIG.  57 

(6)  From  this  result  and  the  known  position  of  the  center  of 
gravity  of  the  triangle,  prove  that  the  center  of  gravity  of  the  segment 
ADB  is  given  by 


3F 


where  F  is  the  area  of  the  segment. 


CENTER   OF  GRAVITY  55 

SUGGESTION.     In  (a)  use  polar  coordinates.     The  element  of  area 
is  pdOdp  and 


X  = 


Problem  50.  Show  that  the  center  of  gravity  of  any  pyramid  is 
in  a  plane  parallel  to  the  base  which  cuts  the  altitude  at  f  the  dis- 
tance from  vertex  to  base. 

Problem  51.  Show  that  the  center  of  gravity  of  a  hemispherical 
surface  bisects  the  radius  perpendicular  to  the  plane  of  the  base  of 
the  surface. 

Problem  52.  Show  that  the  center  of  gravity  of  a  spherical  zone 
is  midway  between  the  planes  of  the  bases  of  the  zone. 

Problem  53.  Show  that  the  center  of  gravity  of  the  surface  of  a 
right  circular  cone  is  at  |  the  distance  from  the  vertex  to  the  base. 

Problem  54.  Show  that  the  center  of  gravity  of  the  frustum  of  a 
right  circular  cone,  the  radii  of  the  bases  being  R  and  r  and  the  alti- 
tude H,  is  distant 

2  #r  +  3  r2 


from  the  base  of  radius  R.     Find  also  the  distance  of  the  center  of 
gravity  from  the  base  of  radius  r. 

Problem  55.  A  casting  is  in  the  form  of  a  hollow  cylinder  with 
one  end  closed.  The  thickness  of  the  end  is  f  in.,  the  length  of  the 
casting  is  12  in.,  and  the  radii  of  the  inner  and  outer  surfaces  of 
the  cylinder  are  5|  and  6  in.  Find  the  position  of  the  center  of  grav- 
ity of  the  casting. 

Problem  56.  If  the  above  casting  is  filled  with  material  £  as 
heavy  as  the  material  of  the  casting,  find  the  position  of  the  center  of 
gravity. 

Problem  57.  A  hemispherical  shell  of  inner  and  outer  radii  r 
and  R  rests  upon  a  hollow  cylinder  of  the  same  material  of  height  H 


56 


APPLIED  MECHANICS  FOR   ENGINEERS 


and  inner  and  outer  radii  r  arid  R.     Find  the  distance  of  the  center 
of  gravity  from  the  base  of  the  cylinder. 

Problem  58.  From  a  right  circular  cone  of  height  20  in.  and 
radius  of  base  10  in.  a  cylinder  of  height  10  in.  and  radius  of 
base  3  in.  is  removed,  the  base  and  axis  of  the  cylinder  being  in 
the  base  and  axis  of  the  cone.  Find  the  distance  from  the  base  of 
the  cone  to  the  center  of  gravity  of  the  part  remaining. 

Problem  59.  If  the  cylindrical  portion  of  the  preceding  problem 
is  filled  with  material  n  times  as  heavy  as  the  material  of  the  cone, 
find  the  distance  from  the  base  to  the  center  of  gravity. 

38.  Center  of  Gravity  of  Counterbalance  of  Locomotive  Drive 
Wheel.  —  In  Fig.  58  the  drive  wheel  is  indicated  by  the 
circle  and  the  counterbalance  by  the  portion  inclosed  by 
the  heavy  lines,  the  point  0  is  the  center  of  the  wheel, 
and  a  is  the  angle  subtended  by  the  counterbalance.  The 


FIG.  58 

point  0'  is  the  center  of  the  circle  forming  the  inner 
boundary  of  the  counterbalance,  and  0  is  the  angle  sub- 
tended by  the  counterbalance  at  this  point.  Let  Fl  repre- 
sent the  area  of  the  segment  of  radius  r  and  F2  the  area 
of  the  segment  of  radius  rr  Also  let  xl  represent  the 
distance  of  the  center  of  gravity  of  Fl  from  0,  and  x%  the 


CENTER   OF  GRAVITY  51 

distance  of  the  center  of  gravity  of  F%  from  0'  .     Then, 
from  Problem  49, 

2  a3        ,  2  a3 


But  #2,  the  distance  of  the  center  of  gravity  of  F2  from  0, 


Then  the  distance  of  the  center  of  gravity  of  the  counter- 
balance from  0  is 


/-? 

r,  cos  -  —  r  cos  - 


where  J^  =  —  —  ar  cos  |, 

and  >  =       !-arCos 


Problem  60.     If  r  =  3  ft.,  rx  =  10  ft.,  and  a  =120°,  find  the  posi- 
tion of  the  center  of  gravity  of  the  counterbalance  of  Art.  38. 

39.   Graphical  Method  of  Finding  the  Center  of  a  Set  of 
Parallel  Forces  in  One  Plane.—  Let  Pv  P2,  P3,  P4  be  a  set 

of  parallel  forces  acting  at  the  points  Av  Av  A%,  A# 
respectively,  in  one  plane.  Assume  the  forces  to  be  acting 
in  the  direction  shown  in  Fig.  59.  Let  one  of  the  forces, 
as  Pv  be  moved  to  any  point  in  its  line  of  action,  as  Sv 
and  resolved  into  two  components  in  any  two  directions, 
as  8^  l  and  $1?  2.  Extend  the  line  of  one  of  these  compo- 


58 


APPLIED  MECHANICS  FOR  ENGINEERS 


nents,  #L  2,  to  cut  the  line  of  a  second  force,  P2,  and 
resolve  the  second  force  into  two  components,  one  of 
which  is  equal  and  opposite  to  the  component,  S^  2,  of 
the  first  force  on  that  line.  Extend  the  line  of  the  other 
component,  $2i  3,  of  the  second  force,  to  cut  the  third 


FIG.  59 

force  and  resolve  the  third  force  into  components  as  in 
the  case  of  the  second  force.  Proceeding  in  this  way 
until  the  last  force  is  reached  there  remain  then  only  two 
unbalanced  components  of  the  original  forces,  one  of  the 
components,  S^  v  of  the  first  force,  and  one,  S^  5,  of  the 
last  force.  The  resultant  of  the  original  forces  is  there- 
fore the  resultant  of  these  two  unbalanced  components. 
By  Art.  27  the  resultant  of  the  parallel  forces  is  parallel 
to  the  forces.  Hence  the  resultant,  P5,  acts  in  a  line, 
parallel  to  the  original  forces,  through  the  intersection, 
_65,  of  the  unbalanced  components,  $6i  l  and  S^  5. 


CENTER   OF  GRAVITY  59 

The  actual  work  of  locating  the  line  of  action  of  the 
resultant  may  be  shortened  by  the  following  considera- 
tion :  the  triangles  of  forces  at  the  vertices  B-^  B%,  B& 
B±  may  be  placed  with  equal  sides  coinciding,  as  in 
Fig.  59  (5). 

The  sides  representing  the  forces  Pv  P2,  P3,  P4  then 
fall  in  succession  on  a  straight  line,  the  beginning  of  any 
force  falling  at  the  end  of  the  preceding  one.  The  lines 
representing  the  components  S^z,  $2)  3,  etc.,  all  meet  at  a 
common  point  0.  The  rays  from  0  are  then  parallel 
respectively  to  the  sides  of  the  polygon  B^B^B^B^B^ 
This  enables  one  to  draw  the  polygon  B^B^B^B^B^ 
without  constructing  the  separate  triangles  B^D^E^  etc. 
Since  $L  2  and  #5i  1  had  arbitrary  directions  given  them, 
the  point  0  may  be  chosen  arbitrarily,  and  the  polygon 
B^B^B^B^B^  constructed  by  drawing  its  sides  parallel 
to  the  corresponding  rays  from  0.  To  sum  up,  the 
method  is  as  follows  : 

Assume  a  direction  in  which  the  forces  act.  Lay  off  in 
succession  the  forces  on  a  line  parallel  to  their  lines  of 
action.  Join  their  points  of  intersection  with  an  arbitrary 
point  0,  Fig.  60  (6).  Construct  a  polygon  with  vertices 
on  the  lines  of  action  of  the  forces  and  with  sides  parallel 
to  the  corresponding  rays  from  0.  The  intersection  of 
the  sides  of  this  polygon,  B5  in  Fig.  60  (a),  drawn  par- 
allel to  the  rays  from  0  to  the  beginning  and  end  of 
the  line  of  forces,  is  a  point  through  which  the  resultant 
passes.  This  gives  the  line  of  the  resultant  for  the  as- 
sumed direction  of  action  of  the  forces.  The  center  of 
the  forces  is  therefore  on  this  line. 


60 


APPLIED  MECHANICS  FOR  ENGINEERS 


In  the  same  way  another  line  of  action  for  another  as- 
sumed direction  may  be  determined.  The  intersection  of 
these  two  lines  is  the  center  of  the  given  forces  (Art.  28). 


FIG.  60 

In  doing  this  work  the  forces  may  be  laid  off  in  suc- 
cession in  any  order.  It  is  only  necessary  to  make  the 
sides  of  the  polygon  B1B2B3  ...  correspond  to  the 
rays  from  0  to  the  line  of  forces;  i.e.  the  side  of  the 
polygon  connecting  any  two  forces  must  be  parallel  to 
the  ray  from  0  to  the  intersection  of  the  same  two  forces. 

The  polygon  B-^B^B^  ...  is  called  the  equilibrium 
polygon.  A  more  general  discussion  will  be  given  in 
the  chapter  on  non-concurrent  forces  in  one  plane. 

Problem  61.  Parallel  forces  of  8,  12,  16,  and  20  Ib.  act  at 
points  (0,  0),  (1,  3),  (3,  1),  and  (5,  4)  respectively.  Find  graphically 
the  line  parallel  to  the  y-axis  in  which  the  center  of  the  forces  lies. 
Make  the  construction  twice,  laying  off  the  forces  in  different  orders. 
Check  by  the  theorem  of  moments. 

Problem  62.  Find  graphically  the  distance  from  the  x-axis  to 
the  center  of  the  forces  of  the  preceding  problem. 

Problem  63.  Weights  of  1200,  1600,  3000,  and  2000  Ib.  act  on  a 
beam  at  distances  of  0,  5,  8,  and  12  ft.  respectively  from  one  end. 


CENTER   OF  GRAVITY 


61 


\ 


\ 


FIG.  61 


Find  graphically  where  their  resultant  cuts  the  beam.  Check  by 
moments. 

Problem  64.  Locate  graphically  the  center  of  gravity  of  the  area 
of  a  quadrant  of  a  circle. 

SUGGESTION.  Divide  the  area  up  into  a  number,  say  eight,  of 
strips  of  equal  width  (Fig.  61).  The  weights  of  the  strips  are  then 
approximately  proportional  to  the  ordinates  at 
their  middle  points  and  act  in  the  lines  of  these 
ordinates.  Hence  use  these  ordinates,  or  any 
proportional  parts  of  them,  as  forces  and  pro- 
ceed as  in  finding  the  center  of  a  set  of  parallel 
forces. 

Problem  65.     Locate  graphically  the  center 
of  gravity  of  the  arc  of  a  quadrant  of  a  circle. 

Problem  66.     Show  how  to  find  graphically, 

approximately,  the  center  of  gravity  of  any  area  in  one  plane. 
Apply  the  method  to  finding  the  center  of  gravity  of  a  given  area. 
Check  by  cutting  the  area  out  of  cardboard  and  balancing. 

Problem  67.  Find  graphically,  approximately,  the  position  of 
the  center  of  gravity  of  a  hemisphere. 

SUGGESTION.  The  hemisphere  may  be  divided  up  into  a  number 
of  parts  by  parallel  planes  equally  spaced.  The  weights  of  these 
parts  will  then  be  approximately  proportional  to  the  squares  of  the 
ordinates  at  the  middle  of  the  generating  strips  of  area.  Lines  pro- 
portional to  the  squares  of  these  ordinates  may  be  constructed  as 
shown  in  Fig.  62  (&) .  These  lines  may  then  be  used  as  forces  and 
the  graphical  solution  obtained  as  in  the  preceding  problems. 

Problem  68.  Find  graphically  the  position  of  the  center  of  gravity 
of  a  solid  of  revolution  obtained  by  revolving  a  parabola  about  its  axis. 

Problem  69.  By  the  method  of  Problem  67  find  the  position  of 
the  center  of  gravity  of  the  frustum  of  a  cone.  Compare  the  result 
with  that  obtained  from  the  answer  to  Problem  54  by  giving  particular 
values  to  R,  r,  and  H. 


62 


APPLIED  MECHANICS  FOE   ENGINEERS 


£ 

I 

I 

I 

I 
f 


\ 


(a) 


FIG.  62 

40.  Simpson's  Rule.  —  When  the  algebraic  equation  of  a 
curve  is  known,  it  is  expressed  as  y  =/(V),  and  the  area 
between  the  curve  and  either  axis  is  usually  determined 
by  integration.  In  Fig.  63  the  area  ABCD  is  expressed 
by  the  integral 


FIG.  63 


CENTER   OF  GRAVITY 


63 


when  the  curve  represented  by  y  =  f(x)  is  continuous 
between  A  and  B. 

In  many  engineering  problems  the  curve  is  such  that 
its  equation  is  not  known,  so  that  approximate  methods  of 


FIG.  64 

obtaining  the  areas  under  the  curve  must  be  resorted  to. 
One  of  these  methods  of  approximation  is  known  as 
Simpson's  Rule.  Suppose  the  curve  in  question  is  the 
curve  AB  (Fig.  64)  and  it  is  desired 
to  find  the  area  between  the  portion 
AB  and  the  #-axis.  Divide  the  length 
b  —  a  into  an  even  number,  n,  of  equal 
parts  (here  w=10).  Consider  the 
portion  CDEF  and  imagine  it  magni- 
fied as  shown  in  Fig.  65.  Pass  a 
parabolic  arc  through  the  points  (7, 
6r,  .Z),  the  axis  of  the  parabola  being 
parallel  to  the  ^-axis;  then  the  area 
ODEF  is  approximated  by  the  area 
of  the  parabolic  segment  CG-DI  plus 
the  area  of  the  trapezoid  ODEF. 


u.2 


H 
FIG.  65 


64  APPLIED  MECHANICS  FOR  ENGINEERS 


.-.  area  CaVJ3F=±(y2  +  yJEF  +  f  |>3- 

since  the  area  of  the  parabolic  segment  is  |  the  area  of  the 

circumscribing    parallelogram.      Since  EH=  —    — 

71 

this  area  may  be  written 

4  #3  +  #4). 


In  a  similar  way  the  next  two  strips  to  the  right  will  have 

\x 
an  area,  —  Q/4  +  4#54-yu),  and  the  next  two  strips,  an 

o 

area,  -^  (?/6  +  4  ?/7  +  #8),  and  so  on.     Adding  all  these  so 
o 

as  to  get  the  total  area  under  the  portion  of  the  curve  AB, 
we  get 

total  area  =  [y0  +  4(^  +  #3  +  #5  +  y7 


or  in  general  for  n  divisions, 
total  area  =  —  —  [y0  +  4(yx  + 


and  this  is  Simpson's  formula  for  determining  approxi- 
mately the  area  under  a  curve.  It  is  easy  to  see  that 
the  smaller  Az,  the  closer  the  approximation  will  be. 

41.  Application  of  Simpson's  Rule.  —  Simpson's  Rule  may 
be  made  use  of  in  determining  approximately  not  only 
areas,  but  volumes  and  moments.  For  if  a  volume  be 
divided  up  by  planes  perpendicular  to  the  #-axis,  distant 
A#  apart,  into  elements  of  volume  Av,  then  dv  =  Adx, 
where  A  is  the  area  of  the  cross  section  at  any  value  of  #, 


CENTER   OF  GRAVITY  65 

and  the  volume  from  x  =  a  to  x  =  b  may  be  expressed  as 
CAdx.     Hence  the  volume  may  be  evaluated  by  Simp- 

J  a 

son's  formula,  the  y  of  the  formula  being  replaced  by  A. 

The  sum  of  the  moments  of  these  elements  of  volume 

/»&  /»& 

with  respect  to  the  y-axis  is  I  xdv  or   I  xAdx  and  may  be 

evaluated  by  Simpson's  formula,  replacing  y  by  xA.     On 
account  of  its  use  in  adding  moments  Simpson's  formula 


XI 
1 

x 

y/  j 

s 

1 
1 

! 

1 
1 

1 

1 

1 

1 

FIG.  66 

may  be  employed  in  finding  the  center  of  gravity  of  areas 
or  volumes  bounded  by  lines  or  surfaces  whose  equations 
are  not  known.  Suppose,  for  example,  it  is  desired  to 
know  the  volume  and  position  of  the  center  of  gravity 
of  a  coal  bunker  of  a  ship  as  shown  in  Fig.  66.  The 
bunker  is  80  ft.  long  and  the  areas  in  square  feet  are  as 
follows:  AQ  =  400,  A1  =  700,  A2  =  650,  A3  =  600,  A±  =  400. 
The  distance  between  the  successive  areas  is  20  ft. 
Applying  Simpson's  formula  for  volume, 

volume  =  [A,  +  4(A  +  -*t>  +  2  A  +  ^]-     ' 


66 


APPLIED  MECHANICS  FOR  ENGINEERS 


Summing  the  moments,  we  obtain 
Z«r-      8° 


where  x0  =  0,  x1  =  20,  x2  =  40,  #3  =  60,  x±  =  80.  The  po- 
sition of  the  center  of  gravity  from  the  fore  end  can  now 
be  obtained  from  the  relation 


An  approximate  value  of  x  may  also  be  obtained  from 
the  equation 


-  _  v^x\  +  vgs'g  +  t^  +  v^ 


where 
and 


+ 


,  etc., 


x\  =  10,  aj'a  =  30,  etc. 


volume  = 


Problem  70.  A  reservoir  with  five- 
foot  contour  lines  is  shown  in  Fig.  67. 
Find  the  volume  of  water  and  the  distance 
of  the  center  of  gravity  from  the  surface 
of  the  water,  if  the  areas  of  the  contour 
lines  are  as  follows:  AQ  =  0,  Al  =  100  sq. 
ft.,  A2  =  200  sq.  ft.,  As  =  500  sq.  ft.,  A4  = 
600  sq.  ft.,  Ab  -  1000  sq.  ft.,  A6  =  1500  sq. 
ft.,  A7  =  2000  sq.  ft.,  A8  =  2500  sq.  ft. 
Making  substitutions  in  the  Simpson's 
formula,  it  becomes,  for  the  volume, 


(3)  (8)' 
Summing  the  moments  by  Simpson's  formula,  we  have 

,x  =  ^- 

(3) (8) 


CENTER   OF  GRAVITY 

where  x0  =  0  ft.,  xl  =  5  ft.,  x2  =  10  ft.,  etc.     Then 


67 


Both  numerator  and  denominator  are  computed  by  Simpson's  formula. 

Compute  x  by  means  of  the  equa- 
tion, 


^  +  ^2+^3  +  ••• 

and  compare  with  the  previous  re- 
sult. 

Problem  71.  Compute  x  for  the 
parabolic  area  of  Fig.  50,  by  using 
Simpson's  Rule,  and  compare  the 
result  with  that  obtained  by  integra- 
tion. 


Problem     72.        By    Simpson's 
Rule  find  the  area  and  center  of  gravity  of  the  rail  section  shown  in 
Fig.  68. 

Beginning  at  the  top,  the  horizontal  distances  are  as  follows: 

ua  =  1.24"  UA  =  1.0" 


Wl2  =  4" 
tiu  =  4.08' 
«M  -  4.24' 
M9  =  2.5" 


u7  =  1.18" 


w5  =  1.0" 


u2  =  2.23" 
M!  =  5.5" 
"0=6" 


Problem  73.     Find  the  center  of  gravity  of  the  deck  beam  section 
shown  in  Fig.  69.     Use  the  equation 

-  =  JXi  +  FJ9  +  FJE'I  +  etc. 
Fl  +  Fz  -f  Fs  +  etc. 

dividing  the  area  of  the  section  into  convenient  areas.  Check  the 
result  thus  obtained  with  that  obtained  by  balancing  a  stiff  paper 
model  over  a  knife  edge. 


68 


APPLIED  MECHANICS  FOR  ENGINEERS 


42.  Durand's  Rule.  —  A  method  of  find- 
ing the  area  of  irregular  areas  was  pub- 
lished by  Professor  Durand  in  the  En- 
gineering  News,  Jan.  18,  1894.  The  rule 
states  that  the  total  area  of  an  irregular 
curve  equals 


FIG.  69 


+  1.1  yn_t+  0.4  yn) 

where  a,  5,  n,  and  the  y's  have  the  same 
meaning  as  in  Simpson's  Rule.  The 
number  of  divisions  may  be  even 
or  odd.  The  student  is  advised 
to  make  use  of  this  rule  as  well  as 
Simpson's  Rule  and  compare  the 
results. 


43.  Theorems  of  Pappus  and  Guldinus.  —  Let  F  be  a  plane 
area  and  OX  a  line  in  its  plane  not  cutting  the  area  F 
(Fig.  70).  Let  the 
plane  containing  the 
area  F  be  revolved 
through  the  angle  a 
about  the  line  OX 
as  an  axis.  Let  dF 
be  an  element  of  the 
area  at  a  distance  y 
from  OX.  Then  the 
element  of  volume  FlG<  70 

generated  by  dF  is   approximately  dv  =  yadF,   and  the 


CENTER   OF  GRAVITY  69 

total  volume  generated  by  the  area  F  is 


=  afydF. 

But  the  distance  of  the  center  of  gravity  of  the  area  F 

CydF 

from  OX  is  given  by  y  =J  • 

F 

.-.  CydF  =  Fy,  and 

V=Fay. 

This  may  be  stated  as  a  general  principle  as  follows  : 
The  volume  of  any  solid  of  revolution  is  equal  to  the  area 

of  the  generating  figure    times   the   distance  its  center  of 

gravity  moves. 

Problem  74.  Find  the  volume  of  a  sphere,  radius  r,  by  the  above 
method,  assuming  it  to  be  generated  by  a  semicircular  area  revolving 
about  a  diameter. 

Problem  75.  Assuming  the  volume  of  the  sphere  known,  find  the 
center  of  gravity  of  the  generating  semicircular  area. 

Problem  76.  Find  the  volume  of  a  right  circular  cone,  assuming 
that  the  generating  triangle  has  a  base  r  and  altitude  h. 

Problem  77.  Assuming  the  volume  of  the  cone  known,  find  the 
center  of  gravity  of  the  generating  triangle. 

Problem  78.  The  parabolic  area  of  Problem  40  revolves  about 
the  a>axis  ;  find  the  volume  of  the  resulting  solid. 

Problem  79.  Find  the  volume  of  an  anchor  ring,  if  the  radius  of 
the  generating  figure  is  a  and  the  distance  of  its  center  from  the  axis 
of  revolution  is  r. 


70 


APPLIED  MECHANICS  FOR  ENGINEERS 


Let  the  curve  AB  (Fig. 
71),  of  length  Z,  be  the  gen- 
erating curve  of  a  surface 
of  revolution.  The  area 
of  the  surface  generated  by 
ds  when  revolved  through 
the  angle  a  is  dF—  ayds, 
and  the  area  of  the  whole 

surface  is  F=  a.  I  yds.     The  center  of  gravity  of  this  curve 
AB  is  given  by  the  expression 

§yds 

y~W= 

% 

yds  =  ly,  and 


\fyds. 


This  may  be  stated  as  follows :  The  area  of  any  surface  of 
revolution  is  equal  to  the  length  of  the  generating  curve  times 
the  distance  its  center  of  gravity  moves. 

Problem  80.     Find  the  surface  of  a  sphere,  radius  r,  assuming  the 
generating  line  to  be  a  semicircular  arc. 

Problem  81.     Find  the  center  of  gravity  of  a  quadrant  of  a  circu- 
lar wire,  radius  of  the  circle  r.     Use  results  obtained  ab.ove. 

Problem  82.     Find  the  surface  of  the  paraboloid  in  Problem  78. 


CHAPTER  V 


COUPLES 

44.  Definitions.  —  Two  numerically  equal  forces  acting 

in  parallel  lines  in  opposite    directions   form   a   couple. 

The  distance  between  the  forces 

is  called  the  arm  of  the  couple. 

The  product  of  one  of  the  forces 

and  the  arm,  with  the  proper 

sign    prefixed,     is     called    the 

moment  of  the  couple.     The  sign 

is   plus   if  the    forces   tend   to 

produce  counter-clockwise  rotation  and  negative  if  they 

tend  to  produce  clockwise  rotation.     If  P  is  one  of  the 

forces,  d  the  distance  between 
them,  and  M  the  moment  of  the 
couple,  then  M=  ±  Pd. 

The  unit  moment  is  that  due 
to  unit  forces  at  unit  distance 
apart.  If  P  =  I  Ib.  and  d  =  1 
ft.,  the  moment  is  called  1  Ib.-ft. 


FIG.  72 


FIG.  73 


45.  Moment  of  a  Couple  about 
Any  Point  in  its  Plane.  — Let  0  be 
any  point  in  the  plane  of  the  couple  of  Fig.  73  and  let  a 
be  the  distance  from  0  to  one  of  the  forces,  P',  as  shown. 

71 


APPLIED  MECHANICS  FOR  ENGINEERS 


(The  forces  are  marked  P  and  P'  to  distinguish  one  from 
the  other;  P  =  P'.)  Then  for  the  point  0  the  sum  of  the 
moments  of  the  forces  P  and  P1  with  respect  to  the  point  is 


for  the  position  01  the  sum  of  the  moments  is 

P(d  -  a)  +  P'a,  or  Pd  ; 
for  the  position  02  the  sum  of  the  moments  is 


Hence,  the  sum  of  the  moments  of  the  forces  forming  a 
couple  with  respect  to  any  point  in  the  plane  of  the  couple  is 
equal  to  the  moment  of  the  couple. 

46.   Combination  of  Couples  in  the  Same  Plane.  —  Let  two 

couples  of  moments  M1  and  Mz  act  in  the  same  plane. 

The  forces  forming  the 
couples  may  always  be 
combined  in  pairs,  one 
force  from  each  couple, 
into  two  numerically 
equal,  parallel  forces 
opposite  in  direction, 
R  and  R'  (Fig.  74). 
These  resultant  forces 

therefore  form  a  couple.     (If  they  fall  in  the  same  line, 

the  moment  of  the  couple  is  zero.) 

Let  M  be  the  moment  of  this  resultant  couple.     If  any 

point  in  the  plane  be  chosen,  then  with  .respect  to  this 

point, 

M—  mom  R  +  mom  R'. 


i 


Fl°-  74 


COUPLES 


73 


But  mom  R  =  mom  Pl  -f  mom  P2 

and  mom  R'  =  mom  P/  4-  mom  P2'.     (Art.  22.) 

Therefore,  adding, 
If  =  (mom  Pl  +  mom  P/)  -f  (mom  P2  +  mom  P2') 


Therefore,  two  couples  in  the  same  plane  are  equivalent  to 
a  single  couple  in  their  plane  whose  moment  is  equal  to  the 
algebraic  sum  of  the  moments  of  the  two  given  couples. 

It  follows  at  once  that  any  number  of  couples  in  the 
same  plane  are  equivalent  to  a  single  couple  in  that  plane 
whose  moment  is  equal  to  the  algebraic  sum  of  the 
moments  of  the  given  couples. 

47.  Equivalent  Couples,  (a)  In  the  same  plane.  — Let 
the  forces  P1  and  P^  form  a  couple  whose  moment  is 
M1  =  Pjdj.  Let  any  two 
parallel  lines  AB  and  CD  dis- 
tant d2  apart  cut  the  lines  of 
action  of  Pj  and  P^  in  A 
and  C  respectively.  Move 
the  forces  P1  and  P/  to  A 
and  0  and  at  these  points 
put  in  two  equal  and  oppo- 
site forces,  &  and  $',  in  the 
line  AC,  of  such  value  that 
the  resultant  of  P1  and  # 
falls  in  the  line  AB.  The 
resultants,  P2  and  P2',  of  Pl  and  S  and  of  P/  and  S'  re- 
spectively, are  then  numerically  equal,  parallel,  and  oppo- 
site in  direction.  They  therefore  form  a  couple. 


FIG.  75 


74  APPLIED  MECHANICS  FOB  ENGINEERS 

By  similar  triangles,  Fig.  75, 


Therefore  P2c?2  =  P^. 

Hence,  a  couple  may  be  replaced  by  any  other  couple  in 
its  plane  provided  only  that  the  moment  of  the  second  couple 
is  equal  to  the  moment  of  the  first. 

(b)  In  parallel  planes.  —  Let  P  and  P'  be  the  forces  of 
a  couple  in  the  plane  MN.  From  two  points  A  and  A1  in 
their  lines  of  action  draw  parallel  lines  to  cut  the  plane 


FIG.  76 

M'N',  parallel  to  MN,  in  the  points  B  and  B' .  At  B  and 
B'  put  in  pairs  of  equal  and  opposite  forces  P  and  P' 
parallel  to  the  forces  P  and  P1  in  the  plane  MN. 
The  two  forces  P  and  P  at  A  and  B'  may  be  combined 
into  a  single  force  2P  acting  at  the  middle  of  the 
line  ABf.  Likewise  P'  and  P'  acting  at  A'  and  B 


COUPLES 


75 


may  be  replaced  by  the  force  2P'  acting  at  the  middle 
of  AB.  But  ABA' B'  is  a  parallelogram,  and  the 
diagonals  bisect  ea'ch  other.  Therefore  the  forces  2P 
and  2  P'  act  at  the  same  point,  and,  since  they  are  equal 
and  opposite,  annul  each  other.  There  are  left  then  the 
forces  P  at  B  and  P'  at  Br,  forming  a  couple  in  the  plane 
'M'N' ,  equal  in  moment  to  the  original  couple.  Hence  a 
couple  may  be  moved  to  any  plane  parallel  to  the  plane  in 
which  it  acts. 

To  sum  up  (#)  and  (5)  :  A  couple  acting  on  any  body 
may  be  replaced  by  any  other  couple  of  the  same  moment 
acting  anywhere  in  the  plane  of  the  couple  or  in  any  par- 
allel plane. 

48.  Moment  of  a  Couple  with  Respect  to  Any  Line.  Defi- 
nition. —  The  moment  of  a  couple  with  respect  to  any  line 
is  the  sum  of  the 
moments  of  the 
forces  forming  that 
couple  with  respect 
to  that  line. 

Let  P  and  P' form 
a  couple  in  the  plane 


HN  whose  moment 
is  M  =  Pd,  and  let 
AB  be  any  line  mak- 
ing the  angle  a  with 
the  normal  to  HN.  Pass  a  plane  HT  perpendicular  to  AB 
to  intersect  the  plane  HN  in  HV.  The  couple  may  be 
moved  in  its  plane  until  the  forces  are  parallel  to  the  line  H  V 


FIG.  77 


76 


APPLIED  MECHANICS  FOR  ENGINEERS 


(Art.  47)  without  changing  the  moment  of  the  couple  with 
respect  to  AB.     For  (Fig.  75)  with  respect  to  any  line, 
mom  _P2  =  mom  Pl  -f  mom  S 
mom  PJ  =  mom  PJ  +  mom  Sf.    (Art.  24.) 


and 
Adding, 


mom  P2  +  mom  P2'  =  mom  P1  +  mom 


since  mom  S  +  mom  Sr  =  0. 

The  projections  of  the  forces  P  and  Pr  on  the  plane  HT 
are  equal  to  the  forces  themselves,  and  the  projection  of 
the  distance,  d,  between  them  is  d  cos  a.  The  moment  of 
the  couple  with  respect  to  AB  is  therefore  Pd  cos  «,  or 
Jfcosa  (Arts.  23  and  45). 

49.  Combinations  of  Couples  in  Intersecting  Planes.  —  Let 
the  couples  of  moments  Ml  and  M2  (Fig.  78)  be  replaced 

in  their  planes  by 
equivalent  couples 
having  a  common 
arm,  a,  coinciding 
with  the  line  of 
intersection  of  the 
planes  (Art.  47). 
The  forces  forming 
the  couples  may  then 
be  combined  into  a 
pair  of  numerically 
equal,  parallel,  and 
opposite  forces  R 
and  R'  which  form  a  couple  in  a  plane  containing  the  line 
of  intersection  of  the  two  given  planes. 


FIG.  78 


COUPLES 


77 


If  a  is  the  angle  between  the  planes,  measured  between 
Pl  and  P2  (Fig.  78),  and  Mis  the  moment  of  the  resultant 
couple,  then 

M  =  aR  =  a^/P?  +  P22  + 


+  (aP2)2+  '2 


cos  a 


cos  a 


+  M£  +  2  M^M^  cos  a. 

The  angle,  0,  which  the  plane  of '  the  resultant  couple 
makes  with  the  plane  of  the  couple  of  moment  Mv  meas- 
ured from  Pl  to  R,  is  given  by 

^o  sin  a 


tan<9  = 


Pl  +  P2  cos  a 


or 


tan0 


sn  a 


M!  + 


cos  a 


Conversely,  the  couple  of  moment  M  may  be  regarded  as  re- 
solved into  the  component  couples  of  moments  M1  and  M¥ 

50.  Vector  Representation  of  Couples.  —  All  of  the  prop- 
erties of  couples  derived  in  the  preceding  articles  may  be 
represented  geometrically  by 
regarding  the  couple  as  a 
vector  quantity  as  follows  ; 
A  couple  may  be  represented 
by  a  vector  perpendicular  to 
the  plane  of  the  couple,  the 
length  of  the  vector  represent- 
ing the  moment  of  the  couple. 
It  is  agreed  that  the  vector 
shall  point  from  the  plane  toward  that  side  from  which 
the  rotation  appears  counter-clockwise. 


FIG.  79 


78 


APPLIED  MECHANICS  FOR  ENGINEERS 


Since  the  couple  may  be  moved  anywhere  in  its'  plane 
or  in  a  parallel  plane,  the  vector  may  be  laid  off  from  any 
point.  Thus  the  vector  AB  may  represent  any  one  of 
the  three  couples  shown  in  Fig.  79.* 

The  theorem  of  Art.  48  expressed  in  terms  of  vectors  is  : 
The  projection  of  the  vector  representing  a  couple  upon  any 
straight  line  represents  the  moment  of  the  couple  with  respect 
to  that  line.  The  theorems  of  Arts.  46  and  49  expressed 
in  terms  of  vectors  are  as  follows :  Couples  may  be  com- 
bined by  adding  the  vectors  representing  them;  and  any 
couple  may  be  replaced  by  two  or  more  couples  whose  added 
vectors  give  the  vector  of  the  original  couple. 

51.  Rectangular  Components  of  a  Couple.  —  A  case  of  par- 
ticular interest  coming  under  the  theorem  just  mentioned 


» 


FIG.  80 


is  the  combination  of  couples  in  three  rectangular  planes 
into  a  single  couple  and  the  converse  problem  of  resolving 
a  couple  into  component  couples  in  three  rectangular 

*  The  arrow  barb  is  placed  a  short  distance  from  the  end  of  the  couple 
vector  to  distinguish  it  from  the  force  vector. 


COUPLES 


79 


planes.  Let  the  planes  be  taken  as  coordinate  planes. 
Call  the  moments  of  the  couples  in  the  2/2-plane,  the  xz- 
plane,  and  the  ^-plane,  M#  My,  and  Mz  respectively. 
Representing  the  couples  as  vectors,  the  values  Mx,  My, 
and  Mz  are  laid  off  along  the  a>,  ^-,  and  2-axes  respectively 
(Fig.  80).  The  vector  representing  the  resultant  couple 
is  then  the  diagonal  of  the  rectangular  parallelopiped 
with  these  vectors  as  edges. 

If  the  moment  of  the  resultant  couple  is  M  and  the 
direction  angles  of  the  vector  representing  the  couple  are 
X,  /Lt,  and  v,  then 


cosX 


M" 


Conversely,  if  \,  /*,  v,  and  M  are  given,  the  component 
couples  in  the  yz-,  xz-,  and  ^-planes  have  moments  equal 
respectively  to  Mx  =  Mcos  X,  Mv  =  Mcos  p,  Mz  =  Mcos  v. 

52.  A  Couple  not  Balanced  by  a  Force.  —  A  force  and  a 
couple  cannot  be  in  equilibrium.  Let  P  and  P'  form  a 
couple  and  let  Pl  be  a  force.  No 
matter  where  the  force  Pl  may  act 
the  couple  may  be  moved  in  its  own 
plane  or  in  a  parallel  plane  until  one 
of  its  forces,  say  P,  has  at  least  one 
point  in  common  with  .the  force  Pv 


The    forces  P    and    Pl    then    either 


FIG.  81 


annul    each    other    or    they    have    a 

resultant   passing   through   their   common    point.      This 

resultant  and  P'  cannot  balance  each  other,  for  two  forces 


80  APPLIED  MECHANICS  FOR  ENGINEERS 

can  balance  only  when  they  have  the  same  line  of  action 
and  are  equal  and  opposite.  Hence  the  force  and  the 
couple  cannot  balance  each  other. 

53.  Substitution  of  a  Force  and  a  Couple  for  a  Force.  —  Let 
P  be  any  force  acting  at  a  point  A  and  0  any  other  point 

distant  d  from  the  line  of 
P.  At  0  put  in  two  equal 
and  opposite  forces  P  and 
Pr  parallel  to  the  force  P 
at  A.  Then  P  at  A  and 
P'  at  0  form  a  couple 

FIG.  82  . 

whose   moment   is   Pa,  and 
there  remains  the  force  P  at  0. 

Hence  any  force  may  be  replaced  by  a  force  equal  and 
parallel  to  the  given  force  acting  at  any  desired  point,  and 
a  couple  in  the  plane  of  the  given  force  and  the  point 
whose  moment  is  equal  to  the  moment  of  the  given  force 
with  respect  to  that  point. 

Problem  83.  Replace  a  force  of  20  Ib.  acting  in  a  line  15  in.  from  a 
point  0  by  a  force  acting  at  0  and  a  pair  of  forces  in  parallel  lines  6  in. 
apart  in  a  plane  parallel  to  the  plane  of  the  given  force  and  the  point  0. 

Problem  84.  Parallel  and  opposite  forces  of  50  Ib.  each  act  in 
lines  distant  20  in.  apart  in  a  plane  the  normal  to  which  has  di- 
rection angles  A.  =  60°,  /A  =  50°,  v  =  ?  .  The  normal  passes  through 
the  first  octant,  and  the  moment  of  the  couple  appears  negative  to  an 
observer  in  the  first  octant  if  the  plane  of  the  couple  is  regarded  as 
passing  through  the  origin.  Find  the  moments  of  this  couple  with 
respect  to  the  coordinate  axes. 

Problem  85.  Determine  the  magnitude  and  plane  of  action  of 
the  resultant  couple  of  the  couples  acting  on  the  rectangular  block 
shown  in  Fig.  83. 


COUPLES 


81 


Problem  86.     Forces  of  10,  15,  and  20  Ib.  act  respectively  in  the 
xz-,  xy-,  and  #z-planes,  parallel  respectively  to  the  z-,  x-,  and  y-axes,  at 
distances  respectively   of   15,   12, 
and  10  in.  from  the  origin.     Re- 
place these  forces  by  a  single  force 
acting  at  the  origin  and  a  couple. 
Find  the  magnitude  and  direction 
of   the   force  vector   and   of   the 
couple  vector. 

Problem  87.  Two  forces,  each 
equal  to  10  Ib.,  act  in  a  vertical 
plane  so  as  to  form  a  positive 
couple.  The  distance  between  the 


ft 

20  LBS. 

^x 

^12  LBS.V 
20  LBS.          /    , 

10  LBS. 

i 

10  LBS. 

o 

12" 

X 

V 

/1  2  LBS. 

r/ 


FIG.  83 


forces  is   2  ft.      Another  couple 

whose  moment  is  equal  to  80  Ib.-in. 

acts  in  a  horizontal  plane  and  is  negative.     Required  the  resultant 

couple,  its  plane,  and  direction  of  rotation. 

Problem  88.  A  couple  whose  moment  is  10  Ib.-ft.  acts  in  the  xy- 
plane ;  another  couple  whose  moment  is  —  30  Ib.-ft.  acts  in  the  xz- 
plane  and  another  couple  whose  moment  is  —25  Ib.-ft.  acts  in  the 
yz-plane.  Required  the  amount,  direction,  and  location  of  the  couple 
that  will  hold  these  couples  in  equilibrium. 

Problem  89.  Show  that  the  moment  of  a  couple  with  respect  to 
an  element  of  a  right  circular  cone,  whose  axis  is  perpendicular  to  the 
plane  of  the  couple,  is  the  same  for  all  elements  of  the  cone. 


CHAPTER   VI 
NON-CONCURRENT  FORCES 

54.  Forces  in  a  Plane.  —  The  most  general  case  of  forces 
in  a  plane  is  that  one  in  which  the  forces  are  non-concur- 
rent and  non-parallel.  We  shall  now  consider  such  a 

T 


FIG.  84 


case.  Let  the  forces  be  Pv  Pv  P8,  P4,  etc.,  as  shown  in 
Fig.  84,  and  let  them  have  the  directions  shown.  Select- 
ing arbitrarily  an  origin  and  a  pair  of  rectangular  axes  in 
the  plane  of  the  forces,  replace  each  force  by  an  equal 


NON-CONCURRENT  FORCES  83 

and  parallel  force  acting  at  the  origin,  and  a  couple 
whose  moment  is  equal  to  the  moment  of  the  force  with 
respect  to  the  origin  (Art.  53).  The  forces  acting  at  the 
origin  may  then  be  replaced  by  a  single  force, 


The  couples,  being  all  in  one  plane,  are  equivalent  to  a 
single  couple  in  that  plane  whose  moment  is 


(Art.  46.) 

Therefore  the  system  of  non-concurrent  forces  in  a  plane 
may  be  reduced  to  a  single  force,  R  =  V  (2  JT)2  +  (2  1^)2, 
acting  at  an  arbitrarily  selected  origin,  and  a  single 
couple  in  the  plane  of  the  forces  whose  moment  is  the 
sum  of  the  moments  of  the  forces  with  respect  to  that 
origin. 

For  equilibrium,  R  =  0  and  ^Pd  ==  0,  or  SJC  =0,  S;r  =  0, 
and  SJPtf  =  0  ;  that  is,  for  equilibrium,  the  sum  of  the 
components  of  the  forces  along  each  of  the  two  axes  is 
zero  and  the  sum  of  the  moments  with  respect  to  any 
point  in  the  plane  is  zero.  Conversely,  if  2JT=  0,  2F=  0, 
and  2Pc?  =  0,  for  a  point  in  the  plane,  the  resultant  force 
R  and  the  resultant  couple  both  vanish  and  the  forces  are 
in  equilibrium. 

55.  Other  Conditions  for  Equilibrium  of  Forces  in  One 
Plane.  —  The  forces  may  be  combined  in  succession  until 
there  is  obtained  either  a  final  resultant,  including  a  zero 
resultant,  or  a  couple  (Arts.  13  and  25).  In  either  case 
the  sum  of  the  moments  of  the  forces  with  respect  to 


84  APPLIED  MECHANICS  FOR   ENGINEERS 

any  point  in  the  plane  is  equal  to  the  moment  of  their 
resultant  force  or  couple  (Arts.  22  and  46).  If  the 
sum  of  the  moments  of  the  forces  with  respect  to  any 
selected  point  is  zero,  the  forces  cannot  be  equivalent 
to  a  couple,  for  the  moment  of  a  couple  is  the  same  for 
all  points  of  the  plane.  The  forces  may,  however,  have 
a  resultant  force.  But  if  in  addition  the  sum  of  the 
moments  of  the  forces  about  two  other  points  of  the  plane 
not  in  the  same  straight  line  with  the  first  point  is  zero, 
the  resultant  must  be  zero  ;  for  either  the  resultant  or 
the  arm  of  the  resultant  must  be  zero,  and  the  arm  can- 
not be  zero  for  three  different  points  not  in  the  same 
straight  line. 

Hence,  if  the  sum  of  the  moments  of  a  set  of  forces  in 
one  plane  with  respect  to  three  points  not  in  the  same 
straight  line  is  zero,  the  forces  are  in  equilibrium. 

It  is  left  as  an  exercise  for  the  student  to  prove  that 
if  the  sum  of  the  rectangular  components  in  one  direction  of 
a  set  of  forces  in  one  plane  is  zero  and  the  sum  of  the 
moments  of  the  forces  with  respect  to  each  of  two  points  in 
the  plane  not  in  a  perpendicular  to  the  given  direction  is 
zero,  the  forces  are  in  equilibrium. 

Problem  90.     Prove  the  statement  of  the  last  paragraph. 

Problem  91.  The  following  forces  in  one  plane  act  upon  a  rigid 
body :  a  force  of  100  Ib.  whose  direction  makes  an  angle  of  45°  with 
the  #-axis,  and  whose  distance  from  the  origin  is  2  ft. ;  also  a  force  of 
50  Ib.  whose  direction  makes  an  angle  of  120°  with  the  :r-axis,  and 
wnose  distance  from  the  origin  is  3  ft. ;  and  a  force  of  500  Ib.  whose  di- 
rection makes  an  angle  of  300°  with  the  a:-axis  and  whose  distance  from 
the  origin  is  6  ft.  The  moments  of  the  given  forces  with  respect  to  the 
origin  are  clockwise.  Find  the  resultant  force  and  the  resultant  couple. 


NON-CONCURRENT  FORCES 


85 


Problem  92.     It  is  required  to  find  the  stress  in  the  members 
AB,  BC,  CD,  and  CE  of  the  bridge  truss  shown  in  Fig.  85. 
NOTE.     The  member 

1   TON  9  TONS 

AB  is  the  member  be- 
tween A  and  B,  the 
member  CD  is  the  mem- 
ber between  C  and  Z>, 
etc.  This  is  a  type  of 
Warren  bridge  truss. 
All  pieces  (members) 


FIG.  85 


4500  LB. 

FIG.  86 


are     pin-connected     so 

that  only  two  forces  act 

on  each  member.     The  members  are,  therefore,  under  simple  tension 

or  compression;  that  is,  in  each  member  the  forces  act  along  the  piece. 
SOLUTION   OF  PROBLEM.     The  reactions  of  the 
supports  are  found  by  considering  all  the  external 
forces    acting    on    the    truss.     Taking    moments 
about  the  left  support,  we  get  the  reaction  at  the 
right  support,  equal  to   4500  Ib.     Summing   the 
vertical  forces  or  taking  moments  about  the  right- 
hand  support,  the  reaction  at  the  left-hand  support 
is  found  to  be  3500  Ib. 
Cutting  the  truss  along  ab  and  putting  in  the  forces  exerted  by  the 

left-hand  portion,  consider  the  right-hand  portion  (see  Fig.  86).     The 

forces  C  and  T  act  along  the  pieces, 

forming    a    system   of    concurring 

forces.     For    equilibrium,    ^X  =  0 

and  *%Y  =  0,  giving  two  equations, 

sufficient    to     determine    the     un- 
knowns C  and  T.     The  forces  in 

the  members  CD  and  CE  may  now 

be  considered  known. 

Cutting  the  truss  along  the  line 

cd   and  putting  in  the  forces  exerted  by  the  remaining  portion  of 

the  truss,  we  have  the  portion  represented  in  Fig.  87.     This  gives  a 


2  TONS 


FIG.  87 


86 


APPLIED  MECHANICS  FOR  ENGINEERS 


system  of  concurring  forces  of  which  C  and  2  tons  are  known,  so 
that  from  the  equations  2,X  =0  and  2,Y  =  0  the  remaining  forces  Cl 
and  TI  may  be  found. 

Problem  93.  In  the  crane  shown  in  Fig.  88  (a)  find  the  forces 
acting  on  the  pins  and  the  tension  in  the  tie  AC.  The  method  of 
cutting  cannot  be  used  in  this  case  since  the  vertical  and  horizontal 
members  are  in  flexure.  Taking  the  horizontal  member  and  consid- 
ering all  of  the  forces  acting  upon  it,  we  have  the  system  of  non  -con- 

curring forces  shown  in  Fig. 
88  (&).  Three  unknowns  are 
involved,  P3,  Pv  and  P2,  and 
these  may  be  determined  by 
three  equations  3X  =  0,  ^Y 
=  0,  and  2Pd  =  0.  It  is  to 
be  remembered  that  the  pin 
pressure  at  E  is  unknown  in 
magnitude  and  direction.  In 
all  such  cases  it  is  usually 
more  convenient  to  resolve 
this  unknown  pressure  into 
its  vertical  and  horizontal 
components,  giving  two 
unknown  forces  in  known 
directions  instead  of  one  un- 
known force  in  an  unknown 
direction.  This  will  be  done 
in  all  problems  given  here. 
3  In  the  present  case  the  two 
forces  Pl  and  P2  are  the  components  of  the  unknown  pin  pressure. 

The  tension  in  the  tie  A  C  may  be  found  by  considering  the  forces 
acting  on  the  whole  crane   and  taking  moments   about  B.     Thus 
=  0  gives,  calling  the  tension  in  the  tie  T, 

T  35  sin  45°  =  8000  (25), 
T  =  8000  (25) 
35  sin  45°' 


NON-CONCURRENT  FORCES 


87 


Problems*.  In 

the  crane  shown 
in  Fig.  89  (a)  find 
the  forces  in  the  ties 
and  the  compres- 
sion in  the  boom. 
The  method  of  cut- 
ting may  be  used 
here  to  determine 
the  compression  T 
and  the  compres- 
sion in  the  boom, 
since  AB  is  not  in 
flexure,  if  we  neg- 
lect its  own  weight. 
Cutting  the  struc- 
ture about  the  point 
A  and  drawing  the 
forces  acting  on 
the  body,  we  have 
the  system  shown  in  Fig.  89 


The  forces  W  may  be  considered  as 
acting  at  the  center  of  the  pulley. 
The  system  of  forces  is  concurring, 
so  that  2X  =  0,  and  ^Y  =  0  are 
sufficient  to  determine  T  and  C. 
T'  may  be  found  by  considering 
the  forces  acting  on  the  whole 
crane  and  taking  moments  about 
the  lowest  point.  What  length  of 
AB  would  make  the  stress,  T7,  ten- 
sion? 

NOTE.  Neglecting  friction,  the 
tension  W  in  the  cord  supporting 
the  weight  is  transmitted  undimin- 
ished  throughout  its  length. 


88 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  95.     Find  the  horizontal  and  vertical  components  of  the 
forces  acting  on  the  pins  of  the  structure  shown  in  Fig.  90. 

SUGGESTION.     First  take  the  vertical  member  and  consider  all  the 

forces  acting  on  it. 

Problem  96.  Find  the 
forces  acting  on  the  pins  of 
the  structure  shown  in  Fig. 
91,  the  weight  of  the  mem- 
bers A  D,  BF,  and  CE  being 
600  lb.,  400  lb.,  and  100  lb.s 
respectively. 

Problem  97.  A  traction 
engine  is  passing  over  a 
bridge,  and  when  it  is  in  the 

position  shown  in  Fig.  92,  one  half  of  the  load  is  carried  by  each  truss. 
The  weight  of  the  engine  is  transmitted  by  the  floor  beams  to  the 
cross  beams,  and  these  are  carried  at  the  pin  connections  of  the  truss. 
Find  the  stress  in  the  members  AB,  BC,  CE,  CD,  and  DF,  for  the 
position  of  the  engine  shown.  The  weight  of  the  engine  is  3  tons. 

NOTE.     The  floor  beams  are  supposed  to  extend  only  from  one 
cross  beam  to  another. 


FIG.  91 


//  / 


FIG.  92 


NON-CONCURRENT  FORCES 


89 


Problem  98.  In  Problem  94,  suppose  the  weight  of  the  boom  to 
be  one  ton  ;  find  the  stresses  T  and  T'  and  the  pin  pressures. 

NOTE.  The  boom  is  now  under  flexure,  so  that  the  method  of  cut- 
ting cannot  be  used. 

Problem  99.  A  dredge  or  steam  shovel,  shown  in  outline  in  Fig. 
93,  has  a  dipper  with  capacity  of  10  tons.  When  the  boom  and  dipper 
are  in  the  position  shown,  find  the  forces  acting  on  AB,  CD,  and  EF. 
The  projection  of  the  point  F  is  6  ft.  from  the  point  E. 


FIG.  93 

SUGGESTION.  Consider  first  all  the  forces  acting  on  CD,  then  all 
the  forces  acting  on  AB. 

NOTE.  The  member  EF  has  been  introduced  as  such  for  the  sake 
of  analysis ;  it  replaces  two  legs,  forming  an  A  frame. 

Problem  100.  Suppose  the  members  of  the  structure  in  Problem 
99  to  have  weights  as.  follows :  AB,  15  tons,  and  CD,  3  tons,  not  in- 
cluding the  10  tons  of  dipper  and  load.  Find  the  forces  as  required 
in  the  preceding  problem. 

Problem  101.  Suppose  the  beam  in  Problem  5  to  be  20  ft.  long 
and  to  have  a  weight  of  2000  Ib. ;  find  the  pin  reaction  and  the  ten- 
sion in  the  tie. 

Problem  102.  Assume  that  the  compression  members  of  the 
Warren  bridge  truss  of  Problem  92  have  each  a  weight  of  500  Ib. ; 
find  the  stresses  in  the  members  BC  and  CE. 


90 


APPLIED  MECHANICS  FOR  ENGINEERS 


56.  Forces  in  Space.  —  First  consider  a  single  force  P 
acting  at  A  (Fig.  94).  Choose  any  point  0  and  rectan- 
gular axes  through  0  as  origin.  Let  the  coordinates  of 


X 


FIG.  94 

A  referred  to  these  axes  be  (z,  ^,  2)  and  let  the  components 
of  P  parallel  to  these  axes  be  JT,  y,  Z.  Construct  a  rec- 
tangular parallelopiped  with  0  and  A  as  opposite  vertices 
and  with  a?,  y,  and  z  as  edges.  Along  these  edges  put  in 
two  pairs  each  of  parallel,  equal,  and  opposite  forces, 
parallel  to  JT,  P",  and  Z  respectively,  one  force  of  each  set 
acting  at  0  in  the  direction  of  the  corresponding  compo- 
nent at  A  (Fig.  94).  It  is  then  evident,  from  an  inspection 
of  the  figure,  that  the  force  P  acting  at  A  is  equivalent 
to  the  components  JT,  Y,  Z  acting  at  0  and  three  couples 
as  follows : 


NON-CONCURRENT  FORCES 


91 


in  the  yz-plaue  a  couple  whose  moment  is  zY—  yZ, 
in  the  #z-plane  a  couple  whose  moment  is  xZ  —  zX, 
in  the  ^-plane  a  couple  whose  moment  is  yX—  xY. 

In  determining  the  sign  of  the  moment  the  observer  is 
supposed    to   look    toward    the 
origin    from    the    positive    end 
of  the  axes  (Fig.  95). 

From  the  above  discussion  of 
a  single  force  it  follows  at  once 
that  any  number  of  forces  in 
space  are  equivalent  to  three 
forces, 

2JT  2F  2Z  r/ 

FIG.  95 
acting   along   rectangular   axes 

at  an  arbitrary  point,  and  three  couples  in  the  yz-,  xz-,  and 
zy-planes  whose  moments  are  respectively 


The  three  forces  acting  at  the  origin  may  be  combined 
into  a  single  resultant, 
R  = 


+  (2  T)2  +  (SZ)2, 


whose  direction  cosines  are 


-K 


(Art.  20.) 


The   couples  may  be  combined  into    a  single  couple  of 
moment 


whose  direction  cosines  are 


_MZ 


=  ^,    cos/.  =  ^,    cosv  =  ^.          (Art.  51.) 
M  M  M 


92  APPLIED  MECHANICS  FOR   ENGINEERS 

For  equilibrium  both  R  =  0  and  M=  0  ;  that  is, 


I 


Since  zY—yZ  is  the  moment  of  the  force  P  with 
respect  to  the  o>axis,  and  hence  Mx  is  the  sum  of  the 
moments  of  all  the  forces  with  respect  to  the  o>axis,  the 
above  conditions  may  be  expressed  in  the  words:  the  sum 
of  the  components  of  the  forces  along  each  of  the  three  arbi- 
trarily chosen  axes  is  zero,  and 
the  sum  of  the  moments  with  re- 
spect to  each  of  these  axes  is  zero 
when  the  forces  are  in  equilibrium. 

Problem  103.  A  vertical  shaft  is 
acted  upon  by  the  belt  pressures  Tl 
and  T2,  the  crank  pin  pressure,  P, 
and  the  reactions  of  the  supports.  (See 
Fig.  96.)  Write  down  the  six  equa- 
tions for  equilibrium. 

NOTE.  The  #-axis  has  been  chosen 
parallel  to  the  force  P,  and  Tl  and  T2 
are  parallel  to  the  a; -axis. 

2X  =  X'  +  X"  -Tl-T2  =  Q, 
^Y  =  Y>  +  F"-P  =  0, 
2Z  =  Z»  -  G-G'  =  Q; 
Mx  =-  Pb  -  Y"l  =  0, 


FIG.  96 


Mz  =  Pa  +  Tjr  -  T2r  =  0. 
From  these  six  equations  six  unknown 
quantities  can   be  found.     If    G,  G', 
FI,  and  T2  are  known,  the  reaction  of  the  supports  and  P  may  be  found. 
Problem  104.     In   Problem   103  suppose  a  =  I',   b  =  8",   c  =  6", 
=  3',  r  =  Q",-G  =  90  lb.,  G'  =  40  lb.,  7\  =  75  lb.,  T2  =  200  lb.;  write 


NON-CONCURRENT  FORCES 


93 


the  equations  for  equilibrium  and  solve  for  P  and  the  reactions  of  the 
supports. 

Problem  105.  Same  as  Problem  104  except  that  7\  and  T2  make 
an  angle  of  30°  (counter-clockwise  when  observed  from  top)  with 
their  directions,  as  shown  in  Fig.  96. 

Problem  106.  A  crane  shown  in  Fig.  97  has  a  boom  45  ft.  long 
and,  a  mast  30  ft.  high.  It  is  loaded  with  20  tons,  and  the  angle 
between  the  boom  and  the  mast  is  45°.  The  legs  each  make  an  angle 
of  30°  with  the  mast  and  are  in  vertical  planes  90°  apart.  Find  the 
pin  pressures  in 
boom  and  mast, 
also  the  stress  in 
the  legs  when 
(a)  the  plane  of 
the  crane  bisects 
the  angle  be- 
tween the  legs 
and  (6)  the  plane 
of  the  crane 

makes  an  angle  of  30°  with  the  vertical  plane  containing  one  of  them. 
If  the  boom  weighs  4000  lb.,  find  the  stress  in  the  legs  when  the  plane 
of  the  crane  bisects  the  angle  between  them.  Assume  that  the  pul- 
leys A  and  B  are  at  the  ends  of  the  boom  and  mast  respectively. 

Problem  107.  Suppose  the  shaft  of  Problem  104  to  be  horizontal, 
find  P  and  the  reactions  of  the  supports.  Assume  y  horizontal  and 
perpendicular  to  the  shaft,  and  x  vertical.  The  c.  g.  of  crank  is  6  in. 
from  the  ar-axis. 

57.  Graphical  Method  for  Forces  in  One  Plane.  —  Let  Pv 
P2,  P3,  P4  (Fig.  98)  be  a  set  of  forces  in  one  plane  ;  to 
find  by  graphical  processes  a  force  P5  that  will  balance 
them. 

in  arbi- 


Resolve Pl  into  two  components,  /S^  2  and 
trary  directions.     Extend  the  line  of  S^2  until  it  meets 


t  r 


94 


APPLIED  MECHANICS  FOR  ENGINEERS 


FIG. 


the  line  of  P2.  Resolve  P2  into  two  components,  $li2, 
equal  and  opposite  to  8^^  of  Pv  and  /S^g.  Extend  the 
line  of  /S^g  to  cut  the  line  of  P3  and  proceed  as  before. 

When  the  last  force,  P4, 
is  reached,  there  remain 
unbalanced  only  the 
component,  S^v  of  the 
first  force  Pv  and  S^  5  of 
the  last  force  P4.  The 
force  P5  that  will  bal- 
ance these  components 
will  balance  the  original 
forces.  Hence  P5  is  de- 
termined in  magnitude, 
direction,  and  line  of 
action  as  the  force  that  will  balance  these  two  compo- 
nents (Fig.  98). 

The  force  P5  is  called  the  equilibrant  of  the  forces 
P  .  .  P 

~1          ^4' 

The  polygon  A^A^  •••  A§,  with  vertices  on  the  lines  of 
the  forces,  is  known  as  an  equilibrium  polygon.  If  the 
sides  of  this  polygon  were  replaced  by  weightless  rods,  or 
links,  hinged  at  the  vertices,  the  forces  Pj  •••  P5  would  be 
just  balanced  by  the  tensions  (or  compressions)  in  the 
rods,  and  the  framework  would  retain  its  position.  An 
indefinite  number  of  equilibrium  polygons  may  be  con- 
structed for  a  set  of  balanced  forces. 

A  shorter  method  of  constructing  the  equilibrium  poly- 
gon and  determining  the  force,  P5,  to  balance  the  given 
forces  is  the  following:  One  triangle  from  each  vertex 


NON-CONCUBEENT  FORCES 


95 


FIG.  99 


A-II  •••  A5  may  be  taken  and  their  common  sides  put 
together,  without  changing  their  directions,  to  form  a 
closed  polygon  whose  sides  are  the  vectors  of  the  forces 
laid  off  in  succession  (Fig.  99).  This  polygon  is  called 
the  force  polygon.  From  the  force  polygon  it  is  evident 
that  the  vector  of  the  force  P5,  the 
equilibrant  of  the  given  forces  Pl  ••• 
P±,  is  the  closing  side  of  the  polygon 
formed  by  laying  off  in  succession  the 
vectors  of  the  given  forces,  the  begin- 
ning of  each  vector  being  placed  at  the 
end  of  the  preceding  one.  Having 
laid  off  the  force  polygon  and  thus  de- 
termined the  magnitude  and  direction 
of  the  equilibrant,  its  line  of  action  is  determined  by 
drawing  the  equilibrium  polygon ;  but  this  may  now  be 
done  without  constructing  the  triangles  at  the  vertices 
Av  A2  •  ••.  For  the  lines  (called  rays)  from  the  point  0 
in  the  force  polygon  are  parallel  to  the  corresponding 
sides  of  the  equilibrium  polygon,  and  since  the  directions 
of  $L2  and  $5il  were  arbitrary,  the  point  0  may  be  any 
point  in  the  plane.  Join  0  to  the  vertices  of  the  force 
polygon  and  then  construct  the  equilibrium  polygon  by 
drawing  its  sides  parallel  to  the  corresponding  rays  of  the 
force  polygon ;  i.e.  a  side  of  the  equilibrium  polygon 
terminating  on  two  lines  of  force  must  be  parallel  to  the 
ray  of  the  force  polygon  drawn  to  the  intersection  of 
those  same  two  forces. 

To  sum  up :  To  determine  graphically  the  equilibrant 
of  a  set  of  forces  in  one  plane,  construct  a  polygon  of  the 


96 


APPLIED  MECHANICS  FOR  ENGINEERS 


vectors  of  the  given  forces.  The  closing  side  is  the 
vector  of  the  equilibrant.  Draw  rays  from  any  point  in 
the  plane  to  the  vertices  of  the  force  polygon.  Construct 
the  equilibrium  polygon  by  drawing  lines  parallel  to  the 
corresponding  rays  of  the  force  polygon,  intersecting  on 
the  lines  of  action  of  the  given  forces.  The  intersection 
of  the  two  sides  of  the  equilibrium  polygon  parallel  to 
the  rays  to  the  closing  side  of  the  force  polygon  is  a 
point  on  the  line  of  action  of  the  equilibrant.  The  equi- 
librant is  thus  completely  determined. 

In  lettering  the  rays  of  the  force  polygon  and  the  sides 
of  the  equilibrium  polygon  it  is  sufficient  to  mark  them 
only  with  the  subscripts  of  the  forces  that  they  connect. 
Figure  100  illustrates  this. 


FIG.  100 


When  the  forces  are  in  equilibrium,  as  Pv  P2,  •••  P&  in 
Fig.  100,  both  the  force  polygon  and  the  equilibrium 
polygon  are  closed.  It  is  possible  to  have  the  force  poly- 
gon close  without  having  the  equilibrium  polygon  close;  as, 


NON-CONCURRENT  FORCES 


97 


for  example,  Pv  •••  P4  and  a  force  P5  parallel  and  equal 
to  the  closing  side  of  the  force  polygon  but  not  passing 


2000  LBS. 


FIG.  101 

through  the  remaining  vertex  of  the  equilibrium  polygon 
(Fig.  101).  The  forces  in  this  case  form  a  couple,  for  the 
forces  P-p  •••P4  are  equivalent  to  a  force  equal  and  oppo- 
site to  P5  passing  through  the  vertex  A5  of  the  equilibrium 
polygon. 

Problem  108.     Determine  graphically  the  equilibrant  of  the  four 
forces  of  Fig.  102.     Where 
does  it  cut  the  line  ASt 

Problem      109.         Find 

graphically     the     reactions 
due  to  the  loads  in  Fig.  103. 

SUGGESTION.     The    force 
polygon     here     becomes     a 
straight   line.     Call  the  re- 
actions P4  and  P5,  and  agree  FlQ  102 
that  P4  shall   follow  P3  in 

the  force  polygon.  P5  must  then  close  the  polygon.  The  missing  ray 
to  the  intersection  of  P4  and  P5  is  found  by  drawing  from  0  a  ray  par- 


1000  LBS. 


500  LBS. 


1500  LBS. 


98 


APPLIED  MECHANICS  FOB  ENGINEERS 


allel  to  the  closing  side  (dotted)  of  the  equilibrium  polygon.     Hence 
P4  and  P5  are  determined.    Check  by  moments. 


P2 


1.5  TONS 


FIG.  103 

When  a  truss  is  acted  upon  by  wind  loads,  the  directions 
of  the  supporting  forces  depend  upon  the  manner  in  which 
the  truss  is  attached  to  the  supports.  Some  trusses  are 
pinned  at  one  end  and  rest  on  rollers  at  the  other.  The 
reaction  at  the  roller  end  may  then  be  assumed  to  be 
vertical,  the  pins  taking  all  of  the  horizontal  load.  If  the 
truss  is  fixed  at  both  ends,  the  reactions  may  be  assumed 

to  be  in  parallel  lines,  or 
else  that  the  supports 
take  equal  parts  of  the 

horizontal  load. 
FIG.  104 

Problem  110.     The  beam 

AB  in  Fig.  104  carries  loads  as  shown.     Find  graphically  the  reactions 
at  A  and  B,  given  that  the  reaction  at  B  is  vertical. 


NON-CONCUBEENT  FORCES 


99 


SUGGESTION.     Since  the  point  A  is  the  only  known  point  on  the 
reaction  through  A,  begin  the  equilibrium  polygon  at  A  as  a  vertex. 

Problem  111.     Find  graphically  the   reactions  of  the  truss  of 
Fig.  105,  assuming  that 

the    reactions     are     in  3  TONS. 

parallel  lines. 

Problem  112.  Find 
graphically  the  reac- 
tions of  the  truss  of 
Fig.  105,  assuming  that 
the  left-hand  reaction 
is  vertical. 


'/////A 


V////A 


FIG.  105 


58.  Stresses  in  Frames.  —  Stresses  in  roof  and  bridge 
trusses  are  usually  computed  on  the  supposition  that  the 
members  are  two  force  pieces;  i.e.  are  pin-connected  at 
the  ends  and  have  loads  applied  only  at  the  joints.  The 
stresses  in  the  truss  of  Problem  92  were  computed  on  that 
supposition.  The  graphical  method  of  determining  the 
stresses  is  often  more  easily  and  quickly  applied  than  the 
analytical  method,  except  for  very  simple  cases.  As  an 
illustration  of  the  graphical  method,  the  stresses  in  the 
truss  of  Fig.  106  are  determined. 

It  is  convenient  here  to  use  Bow's  notation.  Represent 
a  force  acting  on  the  truss,  or  a  member  of  the  truss,  by  a 
pair  of  letters,  one  on  each  side  of  the  line  of  the  force  or 
member.  Thus  the  left-hand  load  on  the  truss  is  read 
AB,  the  right-hand  reaction  is  (7Z>,  the  upper  horizontal 
member  of  the  truss  is  BF,  etc.  Determine  the  reactions, 
either  graphically  or  by  moments.  They  are  (7Z>  =  J^, 
DA  =  J^1-.  The  loads  and  reactions  form  a  system  in 
equilibrium.  These  forces  are  laid  off  on  a  vertical  line, 


100 


APPLIED  MECHANICS  FOR  ENGINEERS 


known  as  the  load  line,  in  succession  in  the  order  in  which 
they  are  met  in  going  around  the  outside  of  the  truss,  as 
AB,  BO,  CD,  DA.  The  force  AB  is  represented  on  the 
load  line  by  the  same  letters  in  small  type,  ab,  etc.  At 
any  joint  of  the  truss  three  or  more  forces  act  in  the 


9  TONS. 


I  2  TONS. 


V 


(b) 


FIG.  106 


directions  of  the  forces  and  members  that  meet  at  that 
joint.  These  forces  are  in  equilibrium  and  must  therefore 
form  a  closed  polygon  when  laid  off  in  succession  (Art. 
16).  In  constructing  the  polygons  the  forces  already  laid 
off  on  the  load  line  are  made  use  of.  Also  a  polygon  of 
forces,  all  of  which  are  known  in  direction,  cannot  be 
constructed  if  more  than  two  forces  are  unknown  in 
magnitude.  Hence  the  force  polygon  is  first  drawn  for  a 
joint  where  only  two  forces  are  unknown.  The  determi- 
nation of  the  forces  acting  at  this  point  gives  additional 
known  forces  at  other  points,  for  the  force  in  any  member 
acts  equally  and  in  opposite  directions  on  the  pins  at  its 
ends. 


NON-CONCURRENT  FORCES  101 

Choosing  the  joint  of  the  truss  at  the  left  support,  the 
forces  DA,  AE,  and  ED  must  form  a  closed  triangle. 
Hence  using  the  force  da  on  the  load  line,  complete  the 
force  triangle  dae  by  drawing  through  d  and  a  lines 
parallel  respectively  to  DE  and  AE  to  meet  in  e.  The 
lines  ae  and  ed  represent  the  forces  in  AE  and  ED.  The 
directions  of  the  forces  acting  at  the  joint  are  those 
given  by  passing  around  the  triangle  dae  in  the  order 
d-a-e,  since  the  force  da  acts  from  d  towards  a.  Next  go 
to  the  joint  ABFE  and  construct  the  force  polygon  for 
that  joint,  using  the  known  forces  ea  and  ab,  drawing 
lines  through  e  and  b  parallel  respectively  to  EF  and  BF 
to  intersect  in  f.  The  polygon  abfea  is  then  the  force 
polygon  for  the  joint  ABFE.  The  directions  of  the 
forces  acting  at  the  joint  are  given  by  passing  around  the 
polygon  in  the  order  a-b-f-e-a,  since  the  force  ab  acts  from 
a  toward  b.  The  direction  in  which  the  forces  act  at  the 
joint  are  indicated  by  putting  arrows  on  the  members 
near  the  joint.  Thus  the  member  AE  pushes  against  the 
pins  at  its  ends,  while  ED  pulls  on  the  pins  at  its  ends. 
AE  is  in  compression,  ED  in  tension. 

The  whole  stress  diagram  (Fig.  106  (£))  is  thus  con- 
structed upon  the  load  line.  The  stress  in  any  member  is 
represented  by  the  line  in  the  force  diagram  terminating 
in  the  same  letters  as  those  of  the  member,  as^  represents 
the  force  in  FG-. 

A  check  on  the  correctness  of  the  work  is  that  the  last 
line  of  the  force  diagram  must  be  parallel  to  the  cor- 
responding member  of  the  truss.  More  accurate  results 
are  obtained  by  drawing  the  figure  to  large  scales. 


102 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  113.  In  the  truss  of  Fig»  107  the  apex  is  distant  £  the 
length  of  the  span  above  the  lower  chord.  The  panels  carry  equal 
loads,  half  the  load  of  each  panel  being  transferred  to  the  truss  at  the 


FIG.  107 


pin  at  the  end  of  that  panel.  Calling  the  total  load  unity,  find  graphi- 
cally the  stresses  in  the  members  of  the  truss.  Write  the  values  of 
the  stresses  on  the  figure  of  the  truss  and  indicate  whether  tension 
or  compression. 

Problem  114.  A  wind  load  acts  on  the  truss  of  Fig.  107,  as  shown 
in  Fig.  108.  Assuming  the  reactions  to  be  in  parallel  lines,  find  the 
reactions  and  stresses  in  the  members  graphically,  and  write  down 
the  stresses  as  in  the  preceding  problem. 


\ 


FIG.  108 


SUGGESTION.     In  finding  the  reactions  the  work  will  be  simplified 
by  combining  the  loads  into  one  load. 


NON-CONCURRENT  FORCES 


103 


Problem  115.  Assuming 
the  total  wind  load  to  be  equal 
to  |  of  the  total  vertical  load 
in  the  preceding  problems  find 
the  stresses  when  both  loads 
act  simultaneously. 

Problem  116.  Find  the 
stresses  in  the  truss  of  Fig. 
109,  all  members  of  the  truss  being  of  the  same  length. 

2  TONS.         2  TONS.  Problem  117.    Find  the 

1  TON  stresses  in  the  members  of 
the  cantilever  truss  of  Fig. 
110. 

Problem  118.  Find  the 
stresses  in  the  truss  of  Fig. 
111. 

59.  Method  of  Substi- 
tution. —  It  sometimes 

FlG-  110  happens    that    in    con- 

structing the  force  diagram  a  point  is  reached  where  it 


1  TON. 


3  TONS. 


3  TONS. 


FIG.  Ill 


104 


APPLIED  MECHANICS  FOB  ENGINEERS 


is  not  possible  to  proceed  to  another  joint  at  which 
there  are  only  two  unknown  forces  acting.  The  temporary 
removal  of  two  members  and  the  substitution  of  another 

member  to  hold  the 
truss  from  collapsing 
will  sometimes  en- 
able one  to  continue 
the  construction  of 
the  force  diagram. 
This  is  illustrated  in 
the  case  of  the  Fink 
truss  shown  in  Fig. 
112. 

Starting  at  the  left 
support,  the  force 
polygons  fag,  abhg, 
and  fghk  can  be  con- 
structed. The  poly- 
gons for  the  remain- 
ing joints  in  the  left 
half  of  the  truss  then 
have  three  sides  miss- 
ing and  cannot  be  completed  in  the  usual  way.  If  the 
members  ML  and  MN  be  removed,  and  a  member  MN 
(Fig.  112(6))  be  substituted,  the  truss  would  stand. 
Moreover,  the  stresses  in  DN,  NE,  and  EF  would  not 
be  changed ;  for  if  sections  were  made  across  these 
members  in  the  two  cases,  the  parts  of  the  truss  to  the 
left  of  the  cuts  would  be  acted  upon  by  the  same  forces, 
and  hence  would  require  the  same  forces  in  DN,  NE,  and 


NON-CONCURRENT  FORCES  105 

EF  to  balance  them.  (The  same  is  true  for  sections 
across  BH,  HK,  and  KF.)  With  the  changed  form  of 
the  truss  the  force  polygons  khbcm'  and  m'cdn  can  be  con- 
structed. Having  located  n,  we  may  return  to  the  origi- 
nal truss  and  construct  the  polygon  nmcd,  then  bcmlkh,  etc. 

Problem  119.  The  panels  of  a  Fink  truss  are  equal  and  carry 
equal  loads.  The  upper  chord  is  inclined  30°  to  the  horizontal,  and 
the  lower  chord  is  horizontal.  Calling  the  total  load  unity,  construct 
the  stress  diagram  and  write  the  stresses  in  the  members  on  the  figure 
of  the  truss,  indicating  tension  and  compression. 


CHAPTER  VII 

MOMENT  OF  INERTIA 

60.  Definition    of    Moment    of    Inertia.  —  The   study    of 
many  problems  in  mechanics  brings  to  our  attention  the 

value  of  the  integral  of  the  form   j  y^dF,  where  dF  repre- 

sents an  infinitesimal  area  and  y  is  the  distance  of  that 
element  from  an  axis  of  reference.  The  value  of  this 
integral  taken  over  a  given  area  is  called  the  moment  of 
inertia  of  that  area  with  respect  to  the  given  axis  of  ref- 

erence.    In  like  manner   Cr^dV  and    Cr2dM  taken  to  in- 

clude all  elements  of  a  given  volume  or  a  given  mass  are 
called  respectively  the  moment  of  inertia  of  the  volume, 
and  mass,  with  respect  to  the  given  axis.  In  these  inte- 
grals e?F"and  dM  represent  infinitesimals  of  volume  and 
mass  respectively,  and  r  the  distance  of  the  infinitesimal 
element  from  the  axis  of  reference.  We  shall  designate 
moment  of  inertia  by  the  letter  I.  Thus  we  write  : 


for  area,  mass,  and  volume,  respectively. 

106 


MOMENT  OF  INERTIA  107 

Many  problems  that  confront  the  engineer  involve  in 
their  solution  the  consideration  of  the  moment  of  inertia. 
This  is  the  case  when  the  energy  of  a  rotating  flywheel, 
for  example,  is  being  determined.  The  energy  of  a  rotat- 
ing body  (Art.  147)  is  expressed  as  follows : 

Kinetic  energy  = 

where  J  is  the  moment  of  inertia  with  respect  to  the  axis 
of  rotation  and  co  is  the  angular  velocity.  (See  Art.  125.) 
It  is  seen  that  the  energy  of  rotating  bodies,  having  the 
same  angular  velocity,  or  the  same  speed,  is  directly  pro- 
portional to  their  moments  of  inertia.  The  quantity, 
therefore,  plays  a  very  important  part  in  the  considera- 
tion of  rotating  bodies. 

Also  the  strength  of  a  beam  or  column  depends  upon 
the  moment  of  inertia  of  the  area  of  the  cross  section  of 
the  beam  or  column  with  respect  to  a  line  of  the  section 
through  the  center  of  gravity  of  the  section. 

In  a  later  chapter  a  reason  will  be  seen  for  the  name 
"  moment  of  inertia."  For  the  present  it  may  be  regarded 
as  a  name  arbitrarily  applied  to  a  quantity  frequently 
used  in  the  applications  of  mechanics.  (See  Art.  147.) 

61.  Radius  of  Gyration.  —  The  moment  of  inertia  of  an 
area  involves  area  times  the  square  of  a  distance.  We 

may  write  J=  j  y^dF  =  Fk*,  where  F  is  the  area  and  Jc  is 

a  distance,  at  which,  if  the  area  were  all  situated,  the 
moment  of  inertia  would  be  unchanged.  This  distance 
k  is  called  the  radius  of  gyration.  In  a  similar  way 


108  APPLIED  MECHANICS  FOE  ENGINEERS 

for  a  mass  we  write  :  I—   (y2dM=  Mk2,  and  for  volume 


62.  Units  of  Moment  of  Inertia.  —  The  moment  of  inertia 
of   an  area  with  respect  to  any  axis  may  be  expressed 
as  Fk2.     The  area  involves  square  inches,  and  k2  is  a  dis- 
tance squared.     The  product  is  expressed  as  inches  to  the 
fourth  power.     The  moment  of  inertia  of  a  volume  Vk2 
requires  inches  to  the  fifth  power.     The  moment  of  in- 
ertia of  a  mass  requires  in  addition  to  Vk2  the  factor  ^, 

9 

so  that  pounds  and  feet  per  second  per  second  are  involved. 
This  is  somewhat  more  complicated  since  it  involves  units 
of  weight,  distance,  and  time.  The  presence  of  g  (  =  32. 2) 
in  the  expression  requires  that  all  distances  be  in  feet. 
It  is  customary  to  express  the  moment  of  inertia  of  a  mass 
without  designating  the  units  used,  it  being  understood 
that  feet,  pounds,  and  seconds  were  used. 

63.  Representation  of  Moment  of  Inertia.  —  From  the  defi- 
nition of  moment  of  inertia  it  is  evident  that  an  area  has 
a  different  moment  of  inertia  for  every  line  in  its  plane. 
We  shall  designate  the  moment  of  inertia  with  respect  to 
a  line  through  the  center  of  gravity  by  Ig  with  a  subscript 
to   indicate   the   particular   gravity   line   intended.     For 
example,  Igx  indicates  the  moment  of  inertia  with  respect 
to   a   gravity  axis   parallel   to   a;,    and   Igv  indicates   the 
moment  of  inertia  with  respect  to  a  gravity  axis  parallel 
to  y.     The  moment  of  inertia  with  respect  to  a  line  other 
than  a  gravity  line  will  be  designated  by  J,  the  proper 


MOMENT  OF  INERTIA 


109 


subscript  indicating  the  particular  line.  Similar  sub- 
scripts will  be  used  to  designate  the  corresponding  radii 
oi  gyration.  It  should  be  noted  that  moment  of  inertia 
is  not  a  quantity  involving  direction.  It  has  to  do  only 
with  magnitude  and  is  essentially  positive. 

64.  Moment  of  Inertia.     Parallel  Axes. 


O)  For  Areas.  In  Fig.  113  (a),  let  OX  and  0'Xf  be 
two  parallel  axes  in  the  plane  of  the  area  F.  If  y  is 
the  ordinate  of  an  element  of  area  dF  referred  to  0'  X' 
and  h  the  distance  between  the  axes,  then 


where  ~y  is  the  ordinate  of  the  center  of  gravity  of  the 
area  Preferred  to  the  axis  0'  Xf  (Art.  36). 

If  0'  Xf  passes  through  the  center  of  gravity  of  F,  then 
y  is  zero  and  the  equation  may  be  written 


110  APPLIED  MECHANICS  FOR  ENGINEERS 

where  Ig  is  the  moment  of  inertia  of  F  with  respect  to  a 
gravity  axis  parallel  to  OX.  This  relation  may  be 
expressed  in  words  as, 

The  moment  of  inertia  of  an  area  with  respect  to  any  line 
in  its  plane  is  equal  to  its  moment  of  inertia  with  respect  to 
a  parallel  gravity  axis  plus  the  area  times  the  square  of  the 
distance  between  the  two  axes. 

(6)  For  Solids.  In  Fig.  113  (5),  OX  and  0'  X'  are 
parallel  axes  distant  h  apart.  dM  is  the  mass  of  an  ele- 
ment of  the  body  distant  r  from  OX  and  r1  from  O'X'. 
If  the  coordinates  of  dM  are  (^,  z),  as  shown  in  the  figure, 
then 


Therefore      Ix  =fr*dM  =/(^2  +  2  hy  +  r'*)dM 


If  O'X'  passes  through  the  center  of  gravity  of  the  solid, 
y  =  0  and  the  equation  reduces  to 

IX   =  Ig   +  WM. 

From  the  above  formulae  it  is  evident  that  the  moment 
of  inertia  of  an  area  or  a  solid  about  a  gravity  axis  is  less 
than  that  for  any  parallel  axis. 

65.  Moment  of  Inertia  ;  Inclined  Axis.  —  It  is  often  desir- 
able, when  Ix  and  Iv  are  known,  to  find  the  moment  of  inertia 
with  respect  to  an  axis  w  making  an  angle  a  with  x.  (See 


MOMENT  OF  INERTIA 


111 


Fig.  114.)     Here,  /„  =§v*dF  and  Iv  =J*u?dF.     In  terms 
of  #,  #,  and  a, 

Iw  =  I  ($  cos  a  —  x  sin  a)2dF 

=  Cy2  cos2  adF-2  Cxy  cos  «  sin  a  <iF  -f  f^2  sin2  a  ^^ 
=  cos2  a  I  ^t?^7  —  2  sin  a  cos  a  I  xydF  +  sin2  a  J 
=  Ix  cos2  a  —  sin  2  a  J  xydF  +  Iy  sin2  a. 

In  a  similar  way 

J^  =  Ix  sin2  a  +  2  sin  a  cos  a  J  ^c?^7  +  Iy  cos2  a, 


where  OF  is  perpendicular  to  OW. 

These  are  the  required  formulae  for  obtaining  the  mo- 
ment of  inertia  with  respect  to  inclined  axes.  It  follows 
that 

Iw  +  lv  =  Ix  +  Iy 


112  APPLIED  MECHANICS  FOB  ENGINEERS 

That  is,  the  sum  of  the  moments  of  inertia  of  an  area 
with  respect  to  two  rectangular  axes  in  its  plane  is  the 
same  as  the  sum  of  the  moments  of  inertia  with  respect  to 
any  other  two  rectangular  axes  in  the  same  plane  and 
passing  through  the  same  point.  This  states  that  the 
sum  of  the  moments  of  inertia  for  any  two  rectangular 
axes  through  a  point  is  constant.  It  will  be  seen  in 
Art.  68  that  this  constant  is  the  polar  moment  of  inertia. 

66.  Product  of  Inertia.  —  The   integral  J  xydF  is  called 

a  product  of  inertia,  for  want  of  a  better  name.  In  case 
the  area  has  an  axis  of  symmetry,  either  the  x-  or  ?/-axis 
may  be  taken  along  such  an  axis.  The  product  of  inertia 
then  becomes  zero,  since  if  x  is  the  axis  of  symmetry,  for 
every  +  y  there  is  a  corresponding  —  y.  A  similar  rea- 
soning shows  that  the  product  of  inertia  is  zero  when  y  is 
the  axis  of  symmetry.  In  such  cases 

Iw  =  Ix  COS2  a  +  Iv  sin2  a 

and  Iv  =  Ix  sin2  a  +  Iy  cos2  a. 

When  J  xydF  is  not  equal  to  zero,  it  is  necessary  to 

select  the  proper  limits  of  integration  and  sum  the  in- 
tegral over  the  area  in  question.  This  is  illustrated  in 
Art.  76. 

67.  Axes  of  Greatest  and  Least  Moment  of  Inertia.  —  It  is 

often  important  to  know  for  what  axis  through  the  center 
of  gravity  the  moment  of  inertia  is  least  or  greatest ;  that 
is,  what  value  of  a  makes  Iw  or  /,,  a  maximum  or  a  mini- 
mum. For  any  area  Ix,  Iy,  and  (  xydF  are  constant  after 


MOMENT  OF  INERTIA  113 

the  x-  and  ?/-axes  have  been  selected.  Using  the  method 
of  the  calculus  for  finding  maxima  and  minima,  we  have, 
putting 

xydF  =  i,  -=*  =  (Iy  —  Ix)  sin  2  a  —  2  i  cos  2  a. 
da 

Equating  the  right-hand  side  to  zero,  the  value  of  a  that 
gives  either  a  maximum  or  minimum  is  seen  to  be  given 
by  the  equation 

tan2a=-^_ 


"-y       -TSB 


or,  what  is  the  same  thing, 

Sin2«= 2i  -and       "  Z.-Z 


It  is  seen  upon  substituting  these  values  of  sin  2  «  and 
cos  2  a  in 

=  2(7,  -  7z)cos  2  a.  +  4  i  sin  2  a 


da2 

that  the  positive  sign  before  the  radical  indicates  a  mini- 
mum and  the  negative  sign  a  maximum  value  for  Iw. 
The  equation  for  tan  2  a  is  satisfied  by  two  values  of  2  a 
differing  by  180°.  These  values  of  2  a  correspond  to  the 
two  signs  in  the  values  for  sin  2  a  and  cos  2  «.  The  values 
of  a  corresponding  to  the  two  signs  therefore  differ  by  90°. 
Hence,  through  any  point  of  the  plane,  there  are  two  axes 
at  right  angles  to  each  other  about  one  of  which  the  moment 
of  inertia  is  a  maximum  and  about  the  other  a  minimum. 
These  axes  are  known  as  the  principal  axes  of  the  area 
through  that  point.  The  axes  through  the  center  of 


114 


APPLIED  MECHANICS  FOB  ENGINEERS 


gravity  of  the  area  for  which  the  moments  of  inertia  was 
greatest  and  least  are  known  as  the  Principal  Axes  of  the 
Area.  This  subject  will  be  further  discussed  in  Art.  76. 

It  is  seen  from  the  above  that  whenj  xydF=  0,  the 
values  of  a.  which  give  maximum  or  minimum  values  for 
/„,  and  Iv  are  0°  and  90°.  This  means  that  the  x-  and  #-axes, 
themselves,  are  the  principal  axes.  Conversely,  if  the  x- 

and  ^-axes  are  principal  axes,  then  ( xydF  =  0.     If  either 

the  x-  or  ^-axis  is  an  axis  of  symmetry,  \  xydF —  0  and 
hence  the  a>  and  «/-axes  are  principal  axes. 

Problem  120.  Derive  the 
formula  for  /„  written  in  Art. 
65. 

Problem  121.  Prove  that 
when  Iw  is  a  maximum,  /„  is  a 
minimum ;  and  that  when  Iw  is 
a  minimum,  Iv  is  a  maximum, 
by  using 

/„,  +  /„  =  constant. 

68.  Polar  Moment  of  In- 
ertia. — The  moment  of  in- 
ertia of  an  area  with  re- 
spect to  aline  perpendicular 
to  its  plane  is  called  the  polar  moment  of  inertia  of  the  area. 
Consider  the  area  represented  in  Fig.  115  and  let  the 
axis  be  perpendicular  to  the  area  at  any  point  0.  Let 
dF  represent  an  infinitesimal  area  and  let  r  be  its  distance 
from  the  axis.  Representing  the  polar  moment  of  inertia 


FIG.  115 


MOMENT  OF  INERTIA 


115 


by  7P,  we  have 


but 
so  that 


or 


!f  = 


That  is,  £Ae  polar  moment  of  inertia  of  an  area  is  equal 
to  the  sum  of  the  moments  of  inertia  of  any  two  rectangular 
axes  through  the  same  point. 

It  has  already  been  shown  that  Ix-\-Iv  —  constant  (see 
Art.  65)  for  any  point  of  an  area. 

69.  Moment  of  Inertia  of  Rectangle.  —  Let  it  be  required 
to  find  the  moment  of  inertia  of  the  rectangle  shown  in 
Fig.  116  (a),  with  respect  to  the  axis  x.  We  may  write 


Since  dF  =  bdy,  this  becomes 


(a} 


FIG.  116 


116  APPLIED  MECHANICS  FOR  ENGINEERS 

To  find  the  moment  of  inertia  with  respect  to  a  grav- 
ity axis  parallel  to  x  we  may  make  use  of  the  formula 
Igx  =  Ix  —  Fd\  from  which  we  have 


The  same  result  may  of  course  be  obtained  by  taking  the 
axes  through  the  center  and  integrating  between  the  limits 

and  -•     From  comparison  we  may  write  the  moment 

of  inertia  with  respect  to  a  gravity  line  perpendicular  to  a?, 
.5,  =  -L*i»  and  *•„-£,    :.,      . 

and  the  polar  moment  of  inertia  for  the  center  of  gravity 


12 

70.  Moment  of  Inertia  of  a  Triangle.  —  It  is  required  to 
find  the  moment  of  inertia  of  the  triangle  shown  in  Fig. 
116  (6)  with  respect  to  the  axis  #,  coinciding  with  the 
base  of  the  triangle.  We  have 

Ix=CfdF,  where  dF=xdy, 

s+h 
»/o 

But  x  =  -(h  —  y),  from  similar  triangles,  giving 
h 

b  Ch  91  1  T& 

h^o  12  6 


MOMENT  OF  INERTIA 


117 


The  moment  of  inertia  with  respect  to  the  horizontal 
gravity  axis  may  now  be  determined. 


J..=  7,-.ra«=,  and  *«,  =      . 

The  same  results  may  of  course  be  obtained  by  direct 
integration. 

Problem  122.  Find  the  moment  of  inertia  of  the  area  of  a 
triangle  with  respect  to  an  axis  through  the  vertex  parallel  to  the 
base. 

Problem  123.  Find  the  polar  moment  of  inertia  of  the  area  of  a 
right  triangle  for  the  center  of  gravity. 

71.  Moment  of  Inertia  of  a  Circular  Area.  —  The  moment 
of  inertia  of  a  circular  area  with  respect  to  a  horizontal 
gravity  axis  re,  as  shown  in  Fig.  117,  may  be  found  as  fol- 


FIG.  117 


lows  :  Igx  =  \y*dF.     Here  it  is  simpler  to  use  polar  coordi- 
nates.    Changing  to  polar  coordinates,  remembering  that 


118  APPLIED   MECHANICS  FOB  ENGINEERS 

y  =  p  sin  0)  and  dF  =  dp(pdd),  the  integral  becomes 


The  corresponding  radius  of    gyration   is  fc^sm—  .      On 

account  of  the  symmetry  of  the  figure  this  is  the  moment 
of  inertia  for  any  line  in  the  plane  through  the  center  of 
gravity.  It  follows  that 


_Z 
°v       4 


and  that 


J    -Z 
"  ""  2 

7    2        ^ 


72.  Moment  of  Inertia  of  Elliptical  Area.  —  Let  it  be 
required  to  find  Igx  and  Igy  of  the  elliptical  area  shown 
in  Fig.  118.  The  equation  of  the  bounding  curve  is 


and 


From  the  equation  of  the  bounding  curve 


so  that 


MOMENT  OF  INERTIA 


119 


FIG.  118 


In  a  similar  way 


-y*dy  =       E,  and  therefore  fca,  =    • 


Since  Ip  =  Igx  +  Igy,  the  polar  moment  of  inertia  with 
respect  to  the  center  is 


,  and  k 


(a2 


It  is  seen  that  when  a  =  b  =  r,  the  equations  obtained 
for  the  elliptical  area  are  the  same  as  those  obtained  for 
the  circular  area,  just  as  they  should  be. 


120 


APPLIED  MECHANICS  FOR  ENGINEERS 


73.  Moment  of  Inertia  of  Angle  Section.  —  When  an  area 
may  be  divided  up  into  a  number  of  triangles,  or  rec- 
tangles, or  other  simple  divisions,  the  moment  of  inertia 
of  the  whole  area  with  respect  to  any  axis  is  often  most 
easily  found  by  taking  the  sum  of  the  moments  of  inertia 
of  the  individual  parts.  This  method  is  often  made  use  of 
in  determining  the  moment  of  inertia  of  such  areas  as  the 
section  of  the  angle  iron,  shown  in  Fig.  119. 

ffV 


J 


t— -; 


\ 


I 


w- 


FIG.  119 


We  shall  now  determine  the  moment  of  inertia  of  this 
section  with  respect  to  the  horizontal  and  vertical  gravity 
axes,  Igx  and  Igv,  and  also  with  respect  to  an  axis  v  (see 
Art.  76),  making  an  angle  a  with  the  axis  x.  Consider 


MOMENT  OF  INERTIA 


121 


.19) 


the  section  divided  into  two  rectangles,  one  5"  xf",  which 
we  may  call  Fv  and  the  other  3f  "  x  f  ",  which  we  may  call 
F2.  The  moment  of  inertia  of  the  section,  with  respect  to 
#,  is  equal  to  the  moment  of  inertia  of  F±  with  respect  to  x 
plus  the  moment  of  inertia  of  F^  with  respect  to  a;,  so  that 
I,,  =  A(6)  (I)3  +  5(f)  (-808)2  +  TV(Y) 
=  7.14  in.  to  the  4th  power.  Similarly 

•k  =  iV(¥XI)3  +  VCDO-Sl)1 
=  12.61  in.  to  the  4th  power. 

NOTE.  The  problem  of  finding  the  moment  of  inertia  of  angle 
sections,  channel  sections,  Z-bar  sections,  and  the  built-up  sections 
shown  in  Figs.  121-125,  is  of  special  interest  and  importance  to 
engineers,  occurring  as  it  does  in  the  computation  of  the  strength  of 
all  beams  and  columns  made  up  of  these  shapes. 


gy 


Problem  124.  Find 
the  moment  of  inertia  of 
the  Z-bar  section  shown 
in  Fig.  120  for  the  grav- 
ity axes  gx  and  gy. 

HINT.  Divide  the  area 
into  three  rectangles. 

Problem  125.  Com- 
pute the  moment  of 
inertia  for  the  channel 
section,  shown  in  Fig.  40,  i  FIG.  120 

Problem  31,  for  the  horizontal  and  vertical  gravity  axes. 

Problem  126.     Required  the  moment  of  inertia  of  the  T-section 
(Fig.  41,  Problem  32),  also  the  moment  of  inertia  of  the  U-section 
(Fig.  42,  Problem  33)  with  respect  to  both  horizontal  and  vertical 
\  gravity  axes. 

A    Problem  127.     The  section  shown  in  Fig.  121  consists  of  a  web 
section  and  4  angles,  as  shown.     Find  the  moment  of  inertia  of  the 


122 


APPLIED  MECHANICS  FOB  ENGINEERS 


whole  section  with  respect  to  the  horizontal  gravity  axis.  Given,  the 
moment  of  inertia  of  an  angle  section  with  respect  to  its  own  gravity 
axis,  g',  is  28.15  in.  to  the  4th  power. 

Problem  128.     Consider  the  section  given  in  Problem  127  to  be 
so  taken  that  it  includes  two  £-inch  rivet  holes,  as  indicated  by  the 


u/: 

*"  i.re"l^l  6"|- 

^ 

j—  ' 

"-1 

t,    »'«/' 

*         \ 

I 

8  --^=^ 

V 

•—y 

^ 

l^ 

r—  f 

La 

IS; 

t"  u  t| 

•9' 


FIG.  121 


FIG.  122 


position  of  the  rivets  in  Fig.  121.  Compute  the  moment  of  inertia  of 
the  whole  section,  when  the  moment  of  inertia  of  the  rivet  holes  is 
deducted.  The  distance  from  the  center  of  the  rivet  hole  to  the  out- 


side  01  tne  angle  sec- 

i 

TF5J""           9'  3   in.      Compare   the 
_i  .    result  with  that  obtained  in 

the  succeeding  problem. 

Problem  129.     The  same 
section  shown  in  Fig.  121  is 
shown  in  Fig.  122  with  two 
~~9     cover  plates.      Find  the  mo- 
inertia  of  the  whole  beam  section  with 
o   its  horizontal  gravity  axis,  now  that 
r  plates  have  been  added.     Allow  for 
vet  holes. 

em  130.     Find  the  moment  of  inertia 
ction  of  a  box  girder,  shown  in  Fig.  123, 
spect   to    its   horizontal   gravity   axis, 
nent  of  inertia  of  one  of  the  angle  sec- 

a 

—  »• 

J 

ment  of 
respect  t 
the  cove 
£-inch  ri 

Uprobl 
of  the  se 

1-,  ^18^  >)      with   re 
Fio.123.               The  moi 

MOMENT  OF  INERTIA 


123 


tions  with  respect  to  its  own  horizontal  gravity  axis,  is  31.92  in.  to 
the  4th  power. 


1 

1 

II 

1 

\ 

\  

1 

1 

^>. 

1 

•9' 


FIG.  124 


Problem  131.     Find  the  moment  of  inertia  of  the  column  section, 
shown  in  Fig.  124,  with  respect  to  the  two  gravity  axes  gx  and  gy. 


T 


Fia.  125 


124 


APPLIED  MECHANICS  FOR  ENGINEERS 


The  column  is  built  up  of  one  central  plate,  two  outside  plates,  and 
four  Z-bars.  The  legs  of  the  Z-bars  are  equal,  and  have  a  length  of 
3£  in.  The  moments  of  inertia  of  each  Z-bar  section  with  respect  to 
its  own  horizontal  and  vertical  gravity  axis  are  42.12  and  15.44  in.  to 
the  4th  power,  respectively. 

i  Problem  132.  Find  the  moment  of  inertia  of  the  section  shown 
in  Fig.  125,  with  respect  to  the  horizontal  and  vertical  gravity  axes 
gx  and  gv.  This  section  is  made  up  of  plates  and  angles.  The 
moment  of  inertia  of  each  angle  section  with  respect  to  both  its  own 
horizontal  and  vertical  gravity  axes  is  28.15  in.  to  the  4th  power. 
The  space  between  the  inner  vertical  plates  is  8f  in. 

74.  Moment  of  Inertia  by  Graphical  Method.  —  It  will 
often  be  necessary  to  find  the  moment  of  inertia  of  a  plane 
section  whose  bounding  curve  is  of  a  complicated  form,  as 


FIG.  126 


in  the  case  when  it  is  necessary  to  compute  the  strength 
of  rails  or  deck  beams.  The  graphical  method  given 
below  may  be  used  for  such  cases. 

Let  F  be  the  area  bounded  by  the  outer  curve  of  Fig. 


MOMENT  OF  INERTIA  125 

126  and  let  OX  be  a  line  with  respect  to  which  it  is  desired 
to  find  the  moment  of  inertia  of  F. 

Draw  a  line  MN  parallel  to  OX  at  a  convenient  dis- 
tance, a,  from  OX.  Let  AB  be  any  chord  of  the  bound- 
ary curve  of  F  parallel  to  OX.  Project  AB  on  MN, 
obtaining  CD.  From  0  and  D  draw  lines  to  any  point 
P  on  OX,  cutting  AB  in  Al  and  Bv  Project  A1B1  on 
MN,  obtaining  O1Dr  Join  Ol  and  Dl  to  P  by  lines 
cutting  AB  in  Az  and  J52.  If  this  be  done  for  all  positions 
of  AB,  a  new  curve  is  obtained,  as  is  shown  in  Fig.  126. 
Let  F"  be  the  area  bounded  by  this  curve. 

Letting  AB  —  x,  AlBl  =  xf,  and  A2B%  =  x"  ,  it  is  seen 
from  similar  triangles  that 

x      a       A    x'      a 

—  =  -  and  —  =  -, 
x'      y  x"      y 

from  which  ~  =  —^     or     x  =  -x". 

x"      yz  y2 

Then,  if  1=  moment  of  inertia  of  F  with  respect  to  OX, 


or  J=«2.F". 

The  area'  F"  may  be  measured  by  a  planinieter  or  by 
one  of  the  methods  previously  explained  (Arts.  40-42). 

If  it  is  desired  to  measure  the  area  of  F"  by  Simpson's 
Rule,  the  area  F  may  first  be  divided  into  an  even  number 
of  strips  of  the  same  width  by  horizontal  lines  and  on  the 
lines  thus  drawn  the  points  of  the  auxiliary  curve  deter- 
mined by  the  method  described  above.  The  area  F11 
will  then  be  ready  for  the  application  of  Simpson's  Rule. 


126  APPLIED  MECHANICS  FOR  ENGINEERS 

If,  in  laying  off  the  area  .F,  1  inch  parallel  to  OX  repre- 
sents m  inches  and  1  inch  perpendicular  to  OX  represents 
n  inches,  then,  measuring  a  in  inches  and  X"1  in  square 
inches,  the  value  Jin  inches4  is  given  by 

I=a*F"mn*. 

It  is  left  for  the  student  to  show  that  the  distance,  #,  of 
the  center  of  gravity  of  F  from  OX  is  given  by 


where  F1  is  the  area  of  the  curve  traced  out  by  Al  and  Bl 
in  Fig.  126. 


\ 


Problem  133.  By  the  above  method  find  the  moment  of  inertia 
of  the  area  of  a  circle  about  a  diameter.  Determine  the  area  of  F" 
by  the  use  of  Simpson's  Rule. 

Problem  134.  By  the  above  method  find  the  moment  of  inertia 
of  a  rectangle  about  one  side.  Take  the  point  P  at  one  of  the  vertices 
of  the  rectangle  and  show  that  the  boundary  curve  of  F"  is  a 
parabola. 

Problem  135.  Find  the  moment  of  inertia  of  the  area  of  the  rail 
section  of  Fig.  68  with  respect  to  the  base  line. 

Using  the  values  of  the  area  and  height  of  the  center  of  gravity 
found  for  the  figure  in  Problem  72,  find  the  moment  of  inertia  of  the 
area  with  respect  to  a  horizontal  gravity  axis. 

75.  Moment  of  Inertia  of  an  Area,  (a)  by  Direct  Addition, 
and  (b)  by  Use  of  Simpson's  Rule. 

(a)  Divide  the  area  into  n  narrow  strips  of  equal 
width  Aa:  parallel  to  the  line  OY  with  respect  to  which 
the  moment  of  inertia  is  desired  (Fig.  127).  If  the 
middle  lengths  of  these  strips  are  respectively  y^y^  •••  yn> 


MOMENT  OF  INERTIA 


and  their  distances  from    OY  respectively 
the  moment  of  inertia  of  the 
area  is  given  approximately 
by  the  formula, 


127 

•»  xm 


or,    since     Ax  = 
approximately, 


b-a 
n 


and, 


Vn 


o 


FIG.  127 

=  yn&x,  the  approximate  formula,  be- 


A1  = 


comes 


1= 


A  more  accurate  result  is  usually  obtained  by  the  use  of 

Simpson's  Rule. 

(5)  Divide  the  area  into  an  even  number  of  strips  of 

width  Ax  parallel  to  the 
line  with  respect  to  which 
the  moment  of  inertia  is 
sought  (Fig.  128). 

Since  /  =   I  x**ydx,  where 

*Ja 

y  is  the  variable  length  of 
the  strips,  it  may  be  eval- 
uated by  Simpson's  Rule 
as  in  Art.  41,  the  quantity 

x*y  taking  the  place  of  y  in  I  ydx.     Hence,  approximately, 

«^a 


FIG.  128 


128  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  136.  By  the  direct  addition  formula  compute  the  mo- 
ment of  inertia  of  a  rectangle  4  in.  by  2  in.  with  respect  to  a  2  in.  side, 
using  (1)  5  divisions,  (2)  10  divisions.  Compare  with  the  exact  value. 

Problem  137.  Solve  the  preceding  problem  by  the  use  of  Simp- 
son's Rule,  using  (1)  6  divisions,  (2)  10  divisions. 

Problem  138.  Find  the  moment  of  inertia  of  the  area  of  the  sec- 
tion of  Fig.  68  with  respect  to  the  base  by  the  method  of  direct 
addition. 

Problem  139.  Solve  the  preceding  problem  by  the  use  of  Simp- 
son's Rule. 

76.  Least  Moment  of  Inertia  of  Area.  —  In  considering 
the  strength  of  columns  and  struts  it  is  necessary  to  know 
the  axis  about  which  the  moment  of  inertia  of  a  cross  sec- 
tion is  a  minimum,  since  bending  will  take  place  about  this 
axis.  It  was  shown  in  Art.  65  that,  if  the  moment  of 
inertia  of  the  area  with  respect  to  two  rectangular  axes  in 
its  plane  is  known,  the  moment  of  inertia  with  respect  to 
any  other  axis,  making  an  angle  a  with  one  of  these,  could 
be  found.  It  was  further  developed  (Art.  67)  that  the 
value  of  a  that  would  render  the  moment  of  inertia  a  min- 
imum was  given  by  the  equation 

2  fxydF 
tan  2  a  =   J _-. 

-Ly       •*•• 

In  case  either  of  the  axes  x  or  y  is  an  axis  of  symmetry,  the 
value  of  «  given  by  this  criterion  is  zero,  so  that,  for  areas 
having  an  axis  of  symmetry,  the  axis  of  least  moment  of 
inertia  is  the  axis  of  symmetry  or  the  one  perpendicular  to  it. 
As  an  illustration  of  the  problem  in  general  let  it  be 
required  to  find  the  least  moment  of  inertia  of  the  angle 


MOMENT  OF  INERTIA  129 

section  shown  in  Fig.  119  with  respect  to  any  axis  in  the 
plane  of  the  area  through  the  center  of  gravity.  Let  v  be 
the  gravity  axis'making  an  angle  a.  with  the  #-axis.  The 
problem  then  is  to  find  such  a  value  of  a  that  Igv  will  be  a 
minimum.  From  Art.  65  we  have 

Igv  =  I0x  cos2  «  —  sin  2  a  \xydxdy  +  Igy  sin2  a. 

In  Art,  73  it  was  found  that  Igx  =  7.14  and  Igv  =  12.61. 

We  proceed  now  to  find  the  value  of  j  xydxdy  for  the 
angle  section.  For  this  purpose,  suppose  the  section  com- 
posed of  two  rectangles,  Fl  (5  in.  x  f  in.),  and  F%  (*£•  in. 
X  f  in.),  and  then  find  the  value  of  the  integral  for  the 
two  rectangles  separately.  Considering  first  the  area  Fv 
and  using  the  double  integration,  we  get 


Ji. 


i.62      _.4&  IM 


In  a  similar  way  for  Fv  we  ha  /e 


J. 


Therefore,    \xydxdy  for  the  whole  area  of  the  angle  sec 
tion  is  5.51  in.  to  the  4th  power.     From  this  we  find 


tan  2«  =  =  2.02. 


130  APPLIED  MECHANICS  FOB  ENGINEERS 

Therefore  2«  =  63°40', 

«=31°50'. 

The  expression  for  Igv  now  becomes 

Igv  =  7.14  cos2  (31°  500-5.51  sin  (63°  40') 

+  12.61  sin2  (31°  50')=  3.72  in.  to  the  4th  power. 
This  gives  the  least  radius  of  gyration, 
^=.84  in. 

Problem  140.  Find  the  least  moment  of  inertia  Igv  and  least 
radius  of  gyration  kgv  of  the  Z-section  shown  in  Fig.  120.  In  this  case 
Igx  =  15.44  in.  to  the  4th  power  and  Iay  =  42.12  in.  to  the  4th  power. 

Ans.     Least  Igv  =  5.61  in.  to  4th  power  and  least  kgv  =  .80  in. 

Problem  141.  An  angle  iron  has  equal  legs.  The  section,  similar 
to  that  in  Fig.  119,  is  8  in.  x  8  in.  with  a  thickness  of  \  in.  Find 
I0x,  I0yy  least  Igv,  and  least  kgv. 

Ans.     Igx  =  Igy  =  48.58  in.  to  the  4th  power, 
Igv  =  19.55  in.  to  the  4th  power. 
kgv  =  1.59  in. 

Problem  142.  Find  the  moment  of  inertia  of  the  column  section, 
shown  in  Fig.  124,  with  respect  to  an  axis  v  making  an  angle  of  30° 
with  gx.  What  value  of  a  makes  Igv  minimum  in  this  case? 

77.  The  Ellipse  of  Inertia.  —  It  is  interesting  to  note,  at 
this  point,  the  relations  between  the  moments  of  inertia 
with  respect  to  all  the  lines,  in  the  plane  of  the  area,  pass- 
ing through  a  point.  We  have  seen  that  for  every  point 
in  an  area  there  is  always  a  pair  of  rectangular  axes  for  one 
of  which  the  moment  of  inertia  is  a  maximum  and  for  the 
other  a  minimum;  that  is,  there  is  always  a  pair  of  prin- 
cipal axes.  The  criterion  for  such  axes  was  found  to  be 


MOMENT  OF  INERTIA  131 

which  means,  since  the  tangent  of  an  angle  may  have  any 
value  from  0  to  infinity,  positive  and  negative,  that  for 
every  point  there  is  always  a  pair  of  axes  such  that 


This  means  that  the  expression  for  Tv  may  always  be 
reduced  to  the  form 

Iv  =  Ix  cos2  a  +  Iy  sin2  a  (Art.  65) 

by  properly  selecting  the  axes  of  reference,  where  now 
Ix  and  Iy  represent  the  principal  moments  of  inertia.  If 
we  divide  through  by  F,  the  equation  becomes 

k2  =  Jc2  Cos2  a  +  k2  sin2  «. 

Let  p  =  —  ,      a  =  —  ,      and  b  =  —  , 

/cv  Ab 

1 


'  1  —  P2  cos2  a  i 


which  is  the  equation  of  an  ellipse  referred  to  the  princi- 
pal axes  of  inertia  as  axes,  the  coordinates  of  any  point 
on  the  ellipse  being  x  =  p  cos  a,  y=  p  sin  a. 

Hence  if  OX  and  OY  are  respectively  the  axes  of 
maximum  and  minimum  moments  of  inertia  of  an  area  for 
a  given  point  0,  there  exists  an  ellipse  with  minor  and 
major  axes  on  OJTand  0  Irrespectively  such  that  the  dis- 
tance from  0  to  the  ellipse  along  any  line  is  the  reciprocal 
of  the  radius  of  gyration  of  the  area  with  respect  to  that 


132 


APPLIED   MECHANICS  FOR  ENGINEERS 


FIG.  129 


line.      This  ellipse  is  called  the  ellipse  of  inertia  of  the 

area  for  the  given  point  (Fig.  129). 

The  ellipse  of  inertia  furnishes 
a  graphical  method  of  finding  the 
moment  of  inertia  of  an  area  for 
any  axis  through  a  point  when  the 
principal  moments  of  inertia  of 
the  area  for  that  point  are  known. 

Problem  143.  Sketch  the  ellipse  of 
inertia  of  the  area  of  a  rectangle  for  (a) 
the  intersection  of  the  diagonals,  (5)  the 
middle  point  of  one  side  of  the  rectangle. 

Problem  144.  Determine  the  princi- 
pal moments  of  inertia  of  the  area  of 
a  rectangle  6  inches  by  10  inches  for  axes  passing  through  a  vertex 
of  the  rectangle.  Sketch  the  ellipse  of  inertia. 

Problem  145.  Determine  the  principal  moments  of  inertia  of  the 
area  of  a  circle  for  a  point  on  its  circumference,  and  sketch  the  ellipse 
of  inertia  of  the  area  for  that 
point. 

78.  Moment  of  Inertia 
of  a  Thin  Plate,  (a)  with 
Respect  to  an  Axis  in  the 
Plate  Parallel  to  its  Faces, 
(6)  with  Respect  to  an  Axis 
Perpendicular  to  its  Faces. 
—  (a)  Suppose  the  plate 
to  be  of  constant  small  thickness  t  and  unit  weight  7 
and  let  x  be  the  distance  of  an  element  of  mass  dM from 

the  axis,  where  dM=^tdF  (Fig.  130). 
9 


FIG.  130 


MOMENT  OF  INERTIA  133 

Then  approximately 

M   2  j  i\/t~    ^y 

9 
But   Cx2dJ7  is  the  moment  of  inertia  of  the  area  of  the 

face  of  the  plate  about  the  given  axis.  Therefore,  ap- 
proximately, the  moment  of  inertia  of  a  thin  plate  with 
reference  to  an  axis  of  the  plate  parallel  to  its  faces  equals 

3-  times  the  moment  of  inertia  of  the  area  of  its  face  with 

9 

reference  to  the  same  axis. 

(b)  Let  r  be  the  distance  of  dM  from  the  axis  perpen- 
dicular to  the  face  of  the  plate  (Fig.  130).  Then  the 
moment  of  inertia  of  the  plate  with  respect  to  the  axis  is 

/*  *"/£  C* 

J  gJ 

Or,  approximately,  the  moment  of  inertia  of  a  thin  plate 
with  reference  to  an  axis  perpendicular  to  the  faces  of  the 

plate  equals  —  times  the  moment  of  in- 

9 

ertia  of  the  face  of  the  plate  with  refer- 
ence to  that  axis. 

79.  Moment  of  Inertia  of  Solid  of  Rev- 
olution.—  Consider  the  moment  of  in- 
ertia of  a  solid  of  revolution  with  respect 
to  its  axis  of  revolution.  Imagine  the 

solid  cut  into  thin  slices,  all  of  the  same 

FIG.  131 

thickness,    by    parallel   planes   perpen- 
dicular to  the  axis  of  revolution  (Fig.  131).     Each  slice 
is  a  circular  disk  of  thickness  dy  and  radius  #,  and  its 


134  APPLIED  MECHANICS  FOR  ENGINEERS 

moment  of  inertia  with  respect  to  the  axis  of  revolution  is 

^-dym?  •  —   (Art.  71).      The  moment  of  inertia  of  the 
9  '  * 

solid  of  revolution  is  the  sum  of  the  moments  of  the  small 
slices,  so  that 

' 


the  limits  of  integration  and  the  relation  between  x  and  y 
depending  upon  the  particular  solid  considered. 

Problem  146.     Prove  that  the  moment  of  inertia  of  a  right  cir- 
cular cylinder  of  radius  r  and  altitude  h  with  respect  to  its  axis  is 
Mr2 
2    ' 

Problem  147.     Prove  that  the  moment  of  inertia  of  a  right  cir- 
cular cone  with  respect  to  its  axis  is  Igh  =  ^  Mr2. 

Problem  148.     Prove  that   the  moment  of  inertia  of  a  sphere 
with  respect  to  a  diameter  is  f  Mr2. 

Problem  149.     The  surface  of  a  spheroid  is  generated  by  re- 

volving the  ellipse  —  +  ^-  =  1  about  the  a>axis.     Prove  that  the  mo- 
ot2     b2 

ment  of  inertia  of  the  solid  inclosed  with  respect  to  the  axis  of 
revolution  is  Igx  =  \  Mb2. 

Problem  150.     Prove  that  the  moment  of  inertia  of  a  rectangular 
parallelepiped  with  edges  a,  &,  and  c,  with  respect  to  a  gravity  line 

M 
12 


parallel  to  the  edge  c  is  Igc  =  ^  (a2  +  62). 


Problem  151.     Prove  that  the  moment  of  inertia  of  a  slender  rod 
of  length  a  with  respect  to  an  axis  perpendicular  to  the  rod  and 

passing  (a)  through  the  center,   (&)  through  the  end,  is   (a)    — — , 


MOMENT  OF  INERTIA 


135 


Problem  152.  An  anchor  ring  is  generated  by  revolving  a  circle 
of  radius  a  about  a  line  in  its  plane  distant  b  from  the  center  of  the 
circle.  Show  that  the.  moment  of  inertia  of  the  mass  of  the  anchor 
ring  with  respect  to  the  axis  of  revolution  is  M(b2  +  -f  a2). 

80.  Moment  of  Inertia  of  a  Body  by  Parallel  Sections.— 
By  dividing  a  body  up  into  thin  plates  by  parallel  planes? 
parallel  to  the  axis  with  respect 
to  which  the  moment  of  inertia 
of  the  body  is  sought,  the  mo- 
ment of  inertia  is  made  to  de- 
pend   upon    the    moments    of 
inertia  of  the  areas  of  the  sec- 
tions and  their  distances  from 
the  given  axis. 

As  an  illustration  consider 
the  problem  of  finding  the  mo- 
ment of  inertia  of  a  right  circu- 

lar cone  with  respect  to  an  axis  through  the  vertex  perpen- 
dicular to  the  axis  of  the  cone  (Fig.  132).  The  moment 
of  inertia  of  the  thin  plate  with  respect  to  the  diameter  of 

its  lower  face  is  approximately 


-  -  (Arts.   78  and 


71).     The  moment  of  inertia  of  this  plate  with  respect  to 
OX  is  therefore 

*"  +  T^dy 


9 


or 


and  the  Ix  for  the  cone  is  then 


136 


APPLIED  MECHANICS  FOR  ENGINEERS 


From  the  figure  x  may  be  evaluated  in  terms  of  y  and 
the  integral  evaluated. 

Problem  153.  Prove  that  the  moment  of  inertia  of  a  right  cir- 
cular cone  with  respect  to  an  axis  through  the  vertex  perpendicular 
to  the  axis  of  the  cone  is  -fa  M(r2  +  4  h2). 

Problem  154.  Using  the  result  of  Problem  153,  find  the  moment 
of  inertia  of  the  cone  about  a  gravity  axis  parallel  to  the  base,  and 

then  about  a  diameter  of  the  base.     Igx  =  —Mir2  -\ — ) . 

Problem  155.  Prove  that  the  moment  of  inertia  of  a  right  circu- 
lar cylinder  with  respect  to  a  diameter  of  the  base  is  M(  r—  -\ — )  • 

Problem  156.  Prove  that  the  moment  of  inertia  of  an  elliptical 
right  cylinder  of  height  h  and  semi-axes  of  base  a  and  b  with  respect 

to  the  diameter  2  a  of  the  base  is  M(—  +  — ),  and  with  respect  to  a 

gravity  line  parallel  to  the  diameter  2  a  is  —(3  b2  +  A2). 

Problem  157.  Show  that  the  moment  of  inertia  of  a  right  circu- 
lar cylinder,  altitude  A,  and  radius  of  base  r,  with  respect  to  a  gravity 

axis  parallel  to  the  base  is  Igx  =  M\*—  +  —  J,  and  find  the  moment 

of  inertia  with  respect  to  an  axis  parallel  to  this  and  at  a  distance  d 
from  the  base. 

Problem  158.  It  is  required  to  find  the  moment  of  inertia  of  the 
cast-iron  disk  flywheel  shown  in  Fig.  133  with  respect  to  its  geomet- 
rical axis. 


FIG.  133 


MOMENT  OF  INERTIA 


137 


HINT.  The  wheel  may  be  regarded  as  made  up  of  three  hollow 
cylinders,  the  moment  of  inertia  of  the  whole  wheel  being  equal  to  the 
sum  of  the  moments  of  inertia  of  the  three  parts.  The  dimensions 
are  as  follows :  diameter  of  wheel  2  ft.,  width  of  rim  and  hub  4  in., 
thickness  of  rim  and  web  2  in.,  thickness  of  hub  1J  in.,  and  diam- 
eter of  shaft  2  in.  All  distances  must  be  in  feet. 

Problem  159.  Find  the  moment  of  inertia  of  the  cast-iron  fly- 
wheel shown  in  Fig.  134  with  respect  to  its  axis  of  rotation.  There 
are  six  elliptical  spokes,  and  these  may  be  regarded  as  of  the  same 
cross  section  throughout  their  entire  length. 


FIG.  134 

Problem  160.  Find  the  mass  and  moment  of  inertia,  with  respect 
to  the  axis,  of  a  right  circular  cylinder  whose  density  varies  directly 
as  the  distance  from  the  axis  and  is  yi  at  the  outside  of  the  cylinder. 

I --Mr* 
~5 

81.  Moment  of  Inertia  of  a  Mass;  Inclined  Axis.  — We 

shall  now  study  the  problem  of  finding  the  moment  of 
inertia  of  a  solid  with  respect  to  an  axis  inclined  to  the 


138 


APPLIED  MECHANICS  FOR  ENGINEERS 


coordinate  axes.  Suppose  the  moments  of  inertia  of  the 
body  with  respect  to  the  three  coordinate  axes  known 
from  the  expressions  : 


and  let  it  be  required  to  find  the  moment  of  inertia  of  the 
body  with  respect  to  any  other  axis   0  V  making  angles 


FIG.  135 


a,  y8,  7  with  the  coordinate  axes.  (See  Fig.  135.)  Let 
dM  equal  the  mass  of  an  infinitesimal  portion  of  the  body 
and  d  its  distance  from  the  axis  0  V. 

Since   r2  =  x*  +  «/2  4-  22,    OA  —  x  cos  a  +  y  cos  /3  -f-  z  cos  7 
and 
cP  =  r2-  OZ2  =  (z2  +  #2  +  z2)  -  (x  cos  «+ y  cos  j3  +z  cos  7>2 


MOMENT  OF  INERTIA  139 

we  may  write 

=  TtC^2  +  y*  +  22)  —  (#  cos  «  4-  y  cos  /3  4-  z  cos  7)2]cOff. 
This  reduces,  since  cos2  a  4-  cos2/9  4-  cos2  7  =  1,  to 

x»  /"• 

4-  I  (z2  4-  ?/2)  cos2  ydM—  2  cos  a  cos  /3  I  xydM 
J  J 

—  2  cos  0  cos  7  (  yzdM  —  2  cos  7  cos  a  J  xzdM, 


or 


Iv  =  Jx  cos2  a  4-  Iy  cos2  /3  4-  Iz  cos2  7  —  2  cos  a  cos  &  I  xydM 
—  2  cos  /3  cos  7  (  yzdM  —  2  cos  7  cos  a  j  xzdM, 

which  gives  the  moment  of  inertia  of  the  body  with 
respect  to  an  inclined  axis  in  terms  of  the  moments  of 
inertia  with  respect  to  the  coordinate  axes  and  the  prod- 
ucts of  inertia  CxydM,  \yzdM,  and  \xzdM. 

82.  Principal  Axes.  —  If    the  three  products   of  inertia 
(xydM,    (yzdM,  and   J  xzdM  are  each  equal  to  zero,  the 
expression  for  Iv  reduces  to  the  form 

Iv  =  Ix  cos2  a  4-  Iv  cos2  /3  +  Iz  cos2  7. 

In  this  case  the  coordinate  axes  x,  y,  and  z  are  called 
the  principal  axes  for  the  point  0  and  the  moments  Ix,  Iy, 
and  Iz,  the  principal  moments  of  inertia  for  that  point. 


140  APPLIED  MECHANICS  FOR  ENGINEERS 

If  the  point  0  is  the  center  of  gravity  of  the  body,  the 
principal  axes  are  called  the  principal  axes  of  the  body.  It 
can  be  shown  that  it  is  always  possible  to  select  the 
coordinate  axes  #,  #,  and  z  so  that  the  products  of  inertia 
given  in  the  expression  for  Iv  will  each  be  zero.  It  fol- 
lows that  for  every  point  of  a  body  there  exists  a  set  of 
rectangular  axes  that  are  principal  axes. 

If  the  #2/-plane  is  a  plane  of  symmetry,  the  products  of 

inertia  \  zydMand  (  zxdMare  both  zero,  for  to  any  term  in 

*LzydM,  as  z^y^dM,  there  corresponds  a  term,  —  z^y^dM, 
equal  in  numerical  value  but  opposite  in  sign.  Hence  if 
two  of  the  coordinate  planes  are  planes  of  symmetry  of  a 
body,  the  three  products  of  inertia  with  respect  to  these 
planes  are  zero,  and  the  coordinate  axes  are  the  principal 
axes  of  the  body  through  the  origin. 

Problem  161.  What  are  the  principal  axes  of  a  sphere  through 
a  point  on  its  surface  ?  What  are  the  values  of  the  principal  mo- 
ments of  inertia  for  that  point  ?  f  Mr2,  I  Mr2,  £  Mr2. 

Problem  162.  Find  the  moment  of  inertia  of  the  ellipsoid  whose 
surface  is  given  by  the  equation 


with  respect  to  the  axes  a,  b,  and  c,  and  with  respect  to  an  inclined 
axis  0V  making  angles  a,  ft,  y  with  a,  b,  and  c,  respectively.  The 
volume  of  an  ellipsoid  is 


,    »  ,  , 

o  o  o  o 

and  Iv  =  Ia  cos2  a  +  Ib  cos2  (3  +  Ic  cos2  y. 

83.  Ellipsoid  of  Inertia.  —  It  is  always   possible   to  re 
duce  the  expression  for  Iv  to  the  form 


MOMENT  OF  INERTIA  141 

'    /„  =  Ix  cos2  a  +  Iy  cos2  j3  +  Iz  cos2  7, 

by  selecting  the  axes  #,  y,  and  z  so  that  the  products  of 
inertia  are  zero. 

Dividing  this  equation  through  by  M,  we  have 

k*  =  kl  cos2  a  +  k  cos2  0  +  1%  cos2  7. 


Let    p  =  —  ,  a  =  —  ,  6  =  —  ,  and  c  =  —  .     Then  the  equa- 
kv          kx          ky  kz 

tion  becomes,  on  multiplying  by  /a2, 

p2  cos2  a     /c?2cos2/3      /32cos27  _  ., 
~~?~         ~P~~  c2 

which  is  the  equation  of  an  ellipsoid  of  semi-axes  a,  6,  and 
c,  on  the  principal  axes,  the  coordinates  of  any  point  on  the 
surface  being 


Hence  for  any  body  there  exists  for  each  point  an 
ellipsoid  such  that  the  distance  from  the  given  point  to 
the  surface  along  any  line  is  the  reciprocal  of  the  radius  of 
gyration  of  the  body  with  respect  to  that  line,  the  axes  of 
the  ellipsoid  coinciding  with  the  principal  axes  for  that 
point. 

Since  one  of  the  semi-axes,  a,  6,  <?,  of  the  ellipsoid  is  a 
maximum  value  and  another  a  minimum  value  of  the  dis- 
tance from  the  center  to  points  on  the  surface,  it  follows 
that  two  of  the  principal  moments  of  inertia  of  a  body 
for  any  point  are  respectively  minimum  and  maximum 
moments  of  inertia  of  the  body  with  respect  to  lines 
through  the  point. 


142 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  163.     Write  the  equation  and  construct  the  inertia  ellip- 
soid for  the  center  of  gravity  of  a  right  circular  cylinder,  altitude  h 

and  radius  r. 

Problem  164.  Construct  the 
inertia  ellipsoid  for  the  center 
of  a  solid  sphere  of  radius  r. 

Problem  165.  Show  that  the 
moment  of  inertia  of  the  seg- 
ment of  the  circle  FI  (Fig.  136) 
with  respect  to  the  axis  OZ  is 


the  moment  of  inertia  of  the 
sector  OBSA,  minus  ^  ahs,  the 
moment  of  inertia  of  the  tri- 
angle OAB,  or 


.  136 


and  the  moment  of  inertia  of  F1  with  respect  to  05  is  *—  /  -     -  sin  a  ), 
the  moment  of  inertia  of  the  sector,  minus  |  haB,  the  moment  of  inertia 

of  the  triangle  A  OB,  or  Ios  =  -la  -  -  sin  a  +  -  sin  2  a  )  . 

o  V          o  o  / 

Problem  166.  Show  that  the  moment  of  inertia  of  the  counter- 
balance (Fig.  58),  with  respect  to  a  line  through  O,  perpendicular  to 
OS,  and  in  the  plane  of  the  wheel,  is  approximately 


where  F2  =  ^  -  ai\  cos^,  00'  -^cos^-r  cos^,  and  t  is  the  thick- 
ness, as  explained  in  Art.  38. 

Problem  167.     Find  the  moment  of  inertia  of  the  counterbalance 
(Fig.  58)^,  with  respect  to  a  line  through  0  perpendicular  to  the  plane 


MOMENT  OF  INERTIA 

of  the  wheel.     It  may  be  written 


143 


84.  Moment  of  Inertia  of  Locomotive  Drive  Wheel.  —  The 
drive  wheel  may  be  represented  as  in  Fig.  137,  and  may  be 
considered  as  made  up  of  a  tire,  rim,  twenty  elliptical 
spokes,  counterbalance,  and  equivalent  weight  on  opposite 
side  of  center,  and  hub. 


FIG.  137 


The  dimensions  of  the  wheel  are  as  follows : 
Tire,  outside  radius  40",  inside  radius  36",  width  5". 
Rim,  outside  radius  36",  inside  radius  34",  width  4-|". 
Hub,  outside  radius  10' ',  inside  radius  4f  ",  thickness  8". 


144  APPLIED  MECHANICS  FOB  ENGINEERS 

Counterbalance,  outside  radius  34",  inside  radius 
V  11.5",  thickness  7|". 

20  spokes,  24"  long,  elliptical  3£"  by  2{". 

Angle  at  center  subtended  by  counterbalance,  «=  94°  40'. 

Radius  of  crank-pin  circle  18". 

7  =  490  Ib.  per  cu.  ft. 

From  the  above  data  the  following  additional  values  are 
computed  (Art.  38): 

Angle  subtended  by  counterbalance  at  center  of  inner 
boundary  circle,  /3  =  30°  20'. 

Mass  of  counterbalance  17.1. 

Distance  of  center  of  gravity  of  counterbalance  from 
center  of  wheel  28.8". 

28  8 
Mass  of  weights  carried  by  crank-pin  =  —  —  x  17.1  =  27.4. 

18 

(Since  moment  of  counterbalance  =  moment  of  crank- 
pin  weights.) 

The  moment  of  inertia  of  the  wheel  with  respect  to  its 
axis  of  rotation  is  first  found.  The  tire,  rim,  and  hub  are 
hollow  cylinders,  and  their  moments  of  inertia  are  com- 
puted from  the  formula 


For  the  counterbalance  the  formula  of  Problem  167  is  used. 
The  spokes  are  regarded  as  elliptical  cylinders  with  the 
short  axis  of  the  ellipse  in  the  plane  of  the  wheel.  The 
moment  of  inertia  of  one  spoke  is  computed  from  the 
formula 


(Arts.  77,72,64.) 

g     ..    . 


MOMENT  OF  INERTIA  145 

and  10  per  cent  is  deducted  for  the  parts  of  the  spokes  in- 
cluded in  the  counterbalance  and  boss. 

The  moment  of  inertia  of  the  weights  at  the  crank-pin 
is  computed  on  the  assumption  that  the  whole  weight  is 
concentrated  at  the  center  of  the  crank -pin. 

The  results  obtained  are  the  following  : 

Tire J0  =  423 

Rim 136 

Hub 7 

Spokes 81 

Counterbalance 120 

Weights  at  crank-pin 62 

Total J0=829 

With  respect  to  a  gravity  axis  OZ  (Fig.  136)  in  the 
plane  of  the  wheel,  when  the  counterbalance  is  in  a  posi- 
tion where  the  line  joining  its  center  of  gravity  to  the 
center  is  perpendicular  to  OZ,  we  get  for  the  moments  of 
inertia  of  the  various  parts : 

Tire Ioz  =  212 

Rim 68 

Hub 4 

Spokes 41 

Counterbalance     .     .     .     ....  103 

Weights  at  crank-pin     .....  62 

Total Joz==490 

Since  the  widths  of  tire  and  rim  are  small  compared  to 
the  diameter  of  the  wheel,  their  moments  of  inertia  are 
closely  approximate  to  the  values  they  would  have  if  the 


146  APPLIED  MECHANICS  FOR   ENGINEERS 

material  of  tire  and  rim  were  in  one  plane,  which  would 
be  one  half  of  the  corresponding  I0.  The  same  is  true, 
less  accurately,  of  spokes  and  hub.  The  values  of  I0  have 
been  divided  by  2  for  the  corresponding  values  of  Ioz  for 
the  four  parts  mentioned.  The  weights  at  the  crank-pin 
center  are  at  the  same  distance  from  OZ  as  from  0,  and 
the  moment  of  inertia  is  therefore  the  same  as  for  0.  The 
moment  of  inertia  of  the  counterbalance  was  computed 
from  the  formula  of  Problem  166. 

Problem  168.  Compute  the  moment  of  inertia  of  a  pair  of 
drivers  and  their  axle  with  respect  to  their  axis  of  rotation.  Use  the 
data  given  above  and  assume  the  axle  as  cylindrical,  the  diameter 
being  9£"  and  the  length  68".  Ans.  1661. 

Problem  169.  Compute  the  moment  of  inertia  of  the  pair  of 
drivers  and  their  axle,  given  in  the  preceding  problem,  with  respect 
to  an  axis  midway  between  the  wheels  and  perpendicular  to  the  axle. 
Consider  the  counterbalance  of  both  wheels  in  such  a  position  as  to 
give  a  maximum  moment  of  inertia  and  the  distance  between  the 
centers  of  the  wheels  60". 

Problem  170.  Find  the  moment  of  inertia  of  two  cast-iron  car 
wheels  and  their  connecting  steel  axle  with  respect  to  (a)  their  axis 
of  rotation,  (ft)  an  axis  midway  between  the  wheels  and  perpendicular 
to  the  axle.  Consider  the  car  wheels  as  composed  of  an  outside 
tread,  a  circular  web,  and  a  hub ;  each  part  may  be  considered  a  hollow 
cylinder  with  the  following  dimensions:  tread,  outside  radius  16", 
inside  radius  14",  width  5|" ;  web,  outside  radius  14",  inside  radius 
5|",  thickness  1.5"  ;  hub,  outside  radius  5£",  inside  radius  2£",  width 
8" ;  axle  (considered  cylindrical),  5"  diameter  and  7'  3"  long.  Dis- 
tance between  centers  of  wheels  60".  According  to  the  assumption 
made  above,  the  flange  has  been  neglected,  the  web  is  considered  a 
hollow  disk,  and  the  axle  of  uniform  diameter  throughout  its 
length. 


MOMENT  OF  INERTIA  147 

Problem  171.  The  value  490  is  the  greatest  value  for  the  moment 
of  inertia  of  a  drive  wheel  with  respect  to  a  gravity  axis  in  its  plane. 
The  least  value  will  be  with  respect  to  an  axis  at  right  angles  to  this 
through  the  centers  of  gravity  of  the  counterbalance  and  wheel.  The 
student  should  compute  this  least  moment  of  inertia. 

Problem  172.  In  Problem  169  the  drivers  have  been  considered 
as  having  their  cranks  in  the  same  plane.  In  practice  they  are  90° 
apart.  Find  the  moment  of  inertia  with  respect  to  the  axis  stated 
when  the  wheels  are  so  placed. 


CHAPTER  VIII 


FLEXIBLE   CORDS.     THE  ARCH.     BENDING  MOMENTS 

85.  Introduction.  —  A  cord  under  tension  due  to  any 
load  may  be  considered  as  a  rigid  body.  In  the  analysis 

of  problems  in  which 
such  cords  are  consid- 
ered, the  method  of 
cutting  or  section  may 
be  used.  Since  the  cord 
is  flexible  (requiring  no 
force  to  bend  it),  it  is 
easy  to  see  that,  no 
matter  what  forces  are 
acting  upon  it,  it  must 
have  at  any  point  the 
direction  of  the  result- 
ant force  at  that  point, 
and  so  must  be  under 
simple  tension.  If  the 
cord  is  curved,  as  is  the  case  where  it  is  wrapped  around 
a  pulley,  the  resultant  force  is  in  the  direction  of  the 
tangent. 

Consider,  as  the  simplest  case,  a  weight  W  suspended 
by  a  cord,  as  shown  in  Fig.  138  (a).  The  forces  acting 
on  Ware  shown  in  (5)  of  the  same  figure.  The  cord  has 
been  considered  cut  and  the  force  T,  acting  vertically 

148 


FIG.  138 


FLEXIBLE  CORDS 


149 


upward,  has  been  used  to  represent  the  tension.  Summa- 
tion of  vertical  forces  =  0  gives  T  =  W.  In  Fig.  138  (<?) 
the  weight  W1  is  supported  by  two  cords.  The  system  of 
forces  acting  on  the  point  0  is  shown  in  (d),  where  T1 
and  T"  represent  the  tensions  in  the  cords  A  and  B 
respectively.  2 X  =  0  and  2  Y  =  0  give 

T1  cos  a  =  T"  cos  ft 
and  T'  sin  a  +  T"  sin  0  =   Wv 

These  two  equations  are  sufficient  to  determine  the 
unknown  tensions  T1  and  T". 

Problem  173.  A  weight  of  500  Ib.  is  attached  to  the  ends  of 
two  cords  of  length  8  ft.  and  6  ft.,  the  other  ends  of  the  cords  being 
attached  to  the  points  A  and  B  respectively,  where  A  is  lower  than 
5,  and  is  distant  9  ft.  horizontally  and  4  ft.  vertically  from  B.  Find 
the  tensions  in  the  cords.  Solve  analytically  and  graphically. 

SUGGESTION.  Make  use  of  the  horizontal  and  vertical  projections 
of  the  cords  to  determine  the  angles  which  the  cords  make  with  the 
horizontal. 

Problem  174.     A  weight  of  275  Ib.  is  knotted  at  a  point  C  to  a 
rope  which  passes  over  two  smooth  pulleys  at  A  and  B  distant  50  ft. 
apart    and  in    the   same   hori- 
zontal line,  carrying  weights  of 
350  Ib.  and  300  Ib.  respectively, 
as  in  Fig.  139.     Find  for  what 
position  of  C  the  weights  will 
be  in  equilibrium.      Solve  ana- 
lytically and  graphically. 

Problem  175.    If,  in  the  pre- 
ceding problem,  B  is  10  ft.  lower  than  A  and  distant  50  ft.  horizon- 
tally from  A,  find  the  position  of  C  for  equilibrium.     Solve  analyti- 
cally and  graphically. 


275  LBS. 

FIG.  139 


150 


APPLIED  MECHANICS  FOR   ENGINEERS 


86.   Several  Suspended    Weights.     Analytical    Method    of 
Solution.  —  If  two  weights  W1  and  W%  are  attached  to  the 

cord,  as  shown  in 
the  case  of  the  cord 
ABOD  (Fig.  140), 
each  portion  is  un- 
der tension.  Consider 
the  cord  cat  at  A 

FIG.  140  and  D  and  represent 

the  tensions    by   T^  and   T%  respectively. 
and  2Y=  0  we  have 


From2Jf=0 


T±  cos  7  =  jT2  cos  a 
and  I7!  sin  7  +  T2  sin  a  =  Wl  +  TT2. 

A  consideration  of  the  forces  acting  at  B,  if  we  call  the 
tension  in  the  portion  BO,  Ty  gives,  when  the  summation 
of  the  x  and  y  forces  are  each  put  equal  to  zero, 

TI  cos  7  =  Ts  cos  /3 
and  TI  sin  7  —  T3  sin  /3  =  IF^. 

In  a  similar  way,  consider  the  forces  acting  on  the  point 
(7,  and  we  have 

Ts  cos  /3  =  T2  cos  a 
and  To  sin  B  +  21  sin  a  =  TK. 


Of  the  six  equations  given  above  only  four  are  independ- 
ent ;  consequently,  of  the  six  quantities  Tv  Tv  T#  a,  /3, 
and  7,  two  must  be  known  in  order  to  determine  the  other 
four,  or  else  additional  conditions  must  be  given  for  deter- 
mining two  more  independent  equations.  Such  condi- 
tions may,  for  example,  be  the  lengths  of  the  cords  and 


FLEXIBLE  CORDS        -  .     151 

the  positions  of  the  two  points  of  support,  A  and  .Z),  of 
Fig.  140. 

In  general,  if  there  are  n  knots  such  as  B  and  C  of 
Fig.  140,  with  the  weights  Wv  W2,  T73,  TF"4,  etc.,  attached, 
it  will  be  possible  to  get  n  +  2  independent  equations. 
These  will  be  sufficient  to  determine  the  tension  in  each 
portion  of  the  cord  and  its  direction,  provided  the  tension 
at  JL,  say,  and  its  direction  are  known.  If  the  weights 
are  close  together,  the  curve  takes  more  nearly  the  form 
of  a  smooth  curve.  Two  special  cases  of  this  kind  are 
discussed  in  this  chapter  in  Art.  97  and  Art.  100. 

Problem  176.  The  points  A  and  D  of  Fig.  140  are  distant  20  ft., 
apart,  and  are  in  the  same  horizontal  line.  Each  of  the  cords  is 
10  ft.  long  and  the  weights  are  each  50  Ib.  Find  the  tensions  in  the 
cords. 

Problem  177.  In  Fig.  140  W^  =  50  Ib.,  W2  =  120  Ib.,  AB  =  10  ft., 
y  —  60°.  D  and  A  are  on  the  same  level,  20  ft.  ap^art.  If  the  tension 
in  AB  equals  80  Ib.,  find  the  tensions  in  CD  and  BC  and  the  lengths 
of  these  cords. 

87.  Graphical  Method  of  Solution. — In  Fig.  141  (i)'  is 
shown  a  cord  carrying  the  weights  Wv  W%,  W&  and  in 
(ii)  the  corresponding  stresses  in  the  cord  and  the  re- 
action at  the  supports.  The  cord  in  this  position  may  be 
thought  of  as  a  rigid  body  in  equilibrium  under  the  action 
of  the  weights  and  the  reactions  of  the  supports.  The 
weights  and  the  reactions  of  the  supports  therefore  form 
a  system  of  forces  in  equilibrium  and  the  relation  found  in 
Art.  57  must  exist  here  between  the  rays  to  the  force 
polygon  and  the  sides  of  the  equilibrium  polygon. 

In  stating  the  problem  for  solution  enough  conditions 
must  be  given  to  determine  the  construction.  If  the 


152 


APPLIED  MECHANICS  FOR  ENGINEERS 


points  of  support  are  given  and  assigned  weights  are  to 
act  in  assigned  vertical  lines,  then  there  exist  an  indefinite 
number  of  solutions.  For  the  tensions  in  the  strings  can 
be  increased  or  decreased  by  shortening  or  lengthening 
them. 

If  the  string  were  cut  at  any  point  and  held  in  position, 
the  part  of  the  string  to  one  side  of  the  cut  would  be  in 


(i) 


FIG.  141 


equilibrium  under  the  action  of  the  forces  acting  upon 
that  part.  The  horizontal  component  of  the  tension  at 
the  point  where  the  cut  is  made  must  then  be  equal  to 
the  horizontal  component  of  the  reaction  at  either  support. 
The  horizontal  component  of  the  tension  of  the  string  is 
therefore  the  same  at  all  points  and  is  equal  to  the  hori- 
zontal component  of  either  reaction. 

If,  now,  in  addition  to  having  given  the  points  of  sup- 
port, the  weights  and  their  lines  of  action,  there  is  also 
given  the  horizontal  component  of  the  tension  in  the 
string,  the  graphical  construction  for  the  shape  of  the 


FLEXIBLE  COEDS  153 

string  and  the  tensions  in  the  segments  can  be  carried  out. 
The  force  polygon  agf---ca  is  laid  off,  the  point  b  not 
being  determined.  Any  point  0  is  chosen  and  the  rays 
oa,  og,  •••  oc  are  drawn.  Horizontal  and  vertical  lines 
are  drawn  through  the  points  of  support  as  action  lines  of 
the  horizontal  and  vertical  components  of  the  reactions  at 
these  points.  Beginning  at  any  convenient  point,  as 
on  AB,  the  equilibrium  polygon  is  drawn  with  sides 
respectively  parallel  to  the  rays  of  the  force  polygon. 
Parallel  to  the  closing  side  (dotted)  of  the  equilibrium 
polygon,  a  ray  (dotted)  is  drawn  from  0  to  ca  determin- 
ing the  point  b.  Therefore  cb  and  ba  are  respectively  the 
vertical  components  of  the  reactions  of  the  right  and  left 
supports  with  the  given  value  of  the  horizontal  compo- 
nent of  the  tension.  The  triangle  bag  is  then  the  triangle  of 
forces  acting  at  the  left  support,  and  bg  therefore  repre- 
sents the  tension  in  the  segment  of  the  cord  attached  there. 
The  segment  of  the  cord  BG-  may  then  be  drawn  parallel 
to  bg  from  the  point  of  support  to  the  vertical  line  in  which 
W1  acts.  In  the  same  way  the  triangles  bgf,  bfe,  and  bed 
are  the  force  triangles  for  the  points  where  the  weights 
are  attached  and  bf,  be,  and  bd  represent  the  tensions  in 
the  segments  BF,  BE,  and  BD  respectively.  They  give 
therefore  the  direction  of  these  segments.  The  shape  of 
the  cord  can  then  be  constructed. 

A  check  on  the  accuracy  will  be  that  the  last  segment 
must  pass  through  the  right  support. 

The  segments  of  the  cord  form  an  equilibrium  polygon, 
or  string  polygon,  for  which  the  corresponding  ray  poly- 
gon has  b  as  a  pole. 


154  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  178.  Weights  of  25,  40,  and  50  Ib.  are  to  be  suspended 
by  a  cord  in  lines  distant  4,  8,  and  11  ft.  respectively  from  the  left 
support.  The  right  support  is  to  be  distant  15  ft.  horizontally  and 
3  ft.  higher  than  the  left.  With  a  horizontal  component  of  tension 
in  the  cord  equal  to  30  Ib.,  find  by  graphical  methods  the  shape  of 
the  cord,  the  lengths  of  the  segments,  the  tensions  in  the  segments, 
and  the  reactions  at  the  support. 

Problem  179.  Ten  equal  weights  are  to  be  hung  at  equal  hori- 
zontal distances  on  a  cord  supported  at  two  points  on  the  same  level. 
The  horizontal  tension  in  the  cord  is  to  equal  five  times  the  value 
of  one  weight.  Draw  the  shape  of  the  string  and  determine  the 
maximum  tension.  Draw  on  the  same  figure  the  shape  of  the  cord 
when  the  horizontal  tension  is  three  times  the  value  of  one  weight. 

88.  Locus  of  the  Pole  of  the  String  Polygon.  —  Let  the  left- 
and  right-hand  vertical  components  of  the  reactions  at 
the  supports  be  Y1  and  Y2  respectively,  If  the  horizontal 
component  of  the  tension,  av  a2,  aB  the  horizontal  distances 
of  the  weights  W^,  W%,  W%  from  the  left  support,  I  the 
horizontal  distance  between  the  supports,  and  h  the  verti- 
cal height  of  the  right  support  above  the  left. 

Taking  moments  about  the  left  support, 

s      h 


*-  I 

Let  r,  =  a1TF1  +  a2T 


Then  F2  =  Y2  4-  -  H. 

The  value  of  Y2  is  the  value  that  Y2  would  have  if  either 
h  or  If  were  zero,  i.e.  if  the  supports  were  on  the  same 
level,  or  if  the  string  were  replaced  by  a  rigid  body  simply 
resting  on  the  supports  without  horizontal  pressure. 


FLEXIBLE  CORDS 


155 


The  value  Y'2  may  be  computed  analytically  and  laid  off 
vertically  from  d  on  the  load  line  gd  (Fig.  142),  or  it  may 
be  located  by  using  the  equilibrium  polygon  for  vertical 


A 
H 


Y, 


6 


B 


F 


wl 


E 


FIG.  142 


forces  only,  assuming  J7"  to  be  zero.  Ir2  is  then  found  by 
increasing  Yf2  by  the  quantity  -  H,  which  is  proportional 

i 

to  IT.  If  through  the  end,  6',  of  Y'2  a  line  is  drawn 
parallel  to  the  line  joining  the  points  of  support,  the  pole, 
6,  of  the  string  polygon  is  found  on  this  line  at  a  horizon- 
tal distance  H  from  the  load  line  gd.  Conversely,  any 
point  on  this  line  may  be  taken  as  a  pole  for  a  string 
polygon,  and  the  corresponding  value  of  J?"is  the  horizon- 
tal distance  from  the  point  chosen  to  the  load  line  gd. 
The  locus  of  the  poles  of  the  string  polygons  is  therefore 
the  straight  line  parallel  to  the  line  joining  the  points  of 
support  and  passing  through  the  points  which  would 
divide  the  load  line  into  the  reactions  if  the  supports  were 
on  the  same  level. 


156 


APPLIED  MECHANICS  FOR  ENGINEERS 


By  varying  the  position  of  the  pole  the  form  of  the 
string  polygon  may  be  changed. 

89.  Pole  Distance  and  Depth  of  String  Polygon.  —  The  depth 
of  the  string  polygon,  measured  from  the  line  joining  the 
points  of  support,  varies  inversely  as  the  pole  distance  from 
the  load  line.  For,  if  d  is  the  depth  of  the  string  polygon 


FIG.  143 

at  a  certain  point  and  D  the  distance  of  the  pole  b  from 
the  load  line  (Fig.  143),  we  have  from  similar  triangles, 

(d  +  z)  :  x  :  :  Y^  :  D, 
and  zi(x-a^)ii  Wl :  D. 

Eliminating  z, 

1 
I) 

Since  av  Yfv  and  W1  are  constants,  and  for  the  given 
position  x  is  constant,  this  equation  shows  that  d  varies 
inversely  as  D.  The  proof  can  be  given  in  like  manner 
for  any  point. 

The  depth  of  the  string  polygon  at  any  point  can  then 
be  made  to  have  any  desired  value  by  a  proper  choice  of 


FLEXIBLE  CORDS  157 

the  pole.  If  with  a  pole  distance  D  the  depth  of  the 
string  polygon  at  a  certain  point  is  d,  and  a  depth  dl  is 
desired,  choose  a  new  pole  distant  Dl  from  the  load  line 

so  that  jDj  =  — ,  and  the  string  polygon  constructed  by 

4j 

the  use  of  this  pole  will  be  the  one  sought. 

Problem  180.  Weights  of  100,  300,  200  Ib.  are  to  be  suspended 
from  a  cord  in  lines  3,  5,  and  8  ft.  respectively  from  the  left  support. 
The  right  support  is  distant  10  ft.  horizontally  and  2  ft.  vertically 
above  the  left  support.  Using  a  scale  of  1  in.  =  2  ft.,  construct  a 
string  polygon  whose  depth  shall  be  4  ft.  at  the  second  load.  From 
the  diagram  scale  off  the  tensions  in  each  part  of  the  cord  and  the 
horizontal  component  of  the  stress  in  the  cord.  Use  a  force  scale  of 
1  in.  =  100  Ib. 

Problem  181.  A  cord  is  suspended  from  two  points  on  the  same 
level  9  ft.  apart.  Eight  weights  of  50  Ib.  each  are  to  be  suspended  at 
equal  intervals  between  the  supports.  The  maximum  tension  in  the 
cord  is  to  be  300  Ib.  Draw  the  shape  of  the  cord.  Find  the  total 
length  of  the  cord  and  the  horizontal  component  of  the  tension. 

Problem  182.  Show  that  when  equal  weights  are  distributed  at 
equal  horizontal  intervals  along  a  cord  suspended  between  two 
supports  on  the  same  level,  the  maximum  tension  in  the  cord  is 
greater  than  half  the  total  load  on  the  cord. 

Problem  183.  A  cord  supported  at  the  ends  on  the  same  level 
carries  a  load  uniformly  distributed  along  the  horizontal.  Draw 
approximately  the  shape  of  the  cord,  given  that  the  tension  of  the 
cord  at  the  lowest  point  is  one  half  the  total  load. 

SUGGESTION.  Divide  the  horizontal  distance  up  into  small  equal 
intervals  and  assume  that  the  weight  in  each  interval  acts  at  the 
center  of  that  interval. 

With  the  vertex  at  the  lowest  point  of  the  string  polygon,  construct 
a  parabola  to  pass  through  the  points  of  support  and  compare  it  with 
the  string  polygon. 


158 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  184.  Make  the  constructions  of  the  above  problem  when 
the  points  of  support  are  not  on  the  same  level. 

Problem  185.  Show  how  to  obtain  the  ratios  of  a  set  of  weights 
to  cause  a  cord  to  hang  in  a  given  string  polygon. 

HINT.     Work  from  the  string  polygon  to  the  force  polygon. 

90.  The  Linked  Arch.  —  Analogous  to  the  problem  of 
finding  the  form  taken  by  a  cord  carrying  loads  at  inter- 
vals is  that  of  finding  the  form  of  a  set  of  connected 
weightless  links  to  sustain  a  given  set  of  loads  in  given 
vertical  lines,  the  links  to  be  in  compression  instead  of 
tension.  Here  again  if  the  horizontal  component  of  the 


r, 


FIG.  144 


thrust  in  any  link  is  given,  the  form  of  the  linkage  can  be 
determined.  The  construction  and  proof  are  analogous  to 
those  for  the  cord  carrying  weights.  (See  Fig.  144.)  The 
links  themselves  lie  in  an  equilibrium  polygon  the  pole  of 
which  is  b.  It  will  be  noted  that  the  pole  of  the  link 
polygon  lies  on  the  opposite  side  of  the  load  line  from  the 
pole  of  the  string  polygon,  and  that  all  poles  of  link 
polygons  lie  on  the  same  straight  line  as  that  on  which 
the  poles  of  the  string  polygons  lie. 


FLEXIBLE  CORDS 


159 


91.  The  Masonry  Voussoir  Arch.  —  In  an  arch  made  of 
links,  constructed  to  carry  a  given  set  of  loads,  any  varia- 
tion of  the  loads,  unless  they  were  all  changed  simulta- 
neously in  the  same  ratio,  would  be  apt  to  cause  the 
collapse  of  the  arch.  In  an  arch  built  of  dressed  stone, 
called  voussoirs,  owing  to  the  size  of  the  bearing  surfaces 
and  the  friction  between  the  surfaces,  an  arch  can  be  built 
to  carry  a  given  set  of  loads  and  allow  for  a  given  varia- 


(c) 


FIG.  145 


tion  in  the  loads  without  endangering  the  structure. 
The  voussoirs  can  be  made  of  such  a  depth  as  to  keep  the 
line  of  the  resultant  pressure  of  two  adjacent  voussoirs 
on  each  other  in  a  certain  portion  of  the  joint.  The 
broken  line  joining  the  points  of  the  resultant  pressures 
of  the  adjacent  arch  stones  on  each  other  is  called  the  line 
of  resistance.  It  is  usually  considered  advisable,  though 
it  is  not  necessary  for  the  stability  of  the  arch,  to  keep  the 
line  of  resistance  within  the  middle  third  of  the  arch.  The 
intensity  of  the  pressure  between  two  arch  stones  is 
assumed  to  vary  uniformly  as  indicated  in  Fig.  145.  In 
(a)  the  resultant  pressure  falls  within  the  middle  third, 
in  (6)  at  the  edge  of  the  middle  third,  and  in  (c)  outside 
the  middle  third.  In  the  latter  case  there  is  a  tendency 


160  APPLIED  MECHANICS  FOB  ENGINEERS 

for  the  joint  to  open.  With  the  line  of  resistance  below 
the  middle  third  there  would  be  a  tendency  to  open  the 
joint  on  the  upper  part  of  the  arch,  and  also  to  unduly  in- 
crease the  pressure  on  the  lower  part  of  the  joint.  Where 
the  line  of  resistance  falls  above  the  middle  third  the  joint 
would  tend  to  open  on  the  lower  side. 

It  is  desired  here  to  present  only  the  elementary  prin- 
ciple of  the  arch.  An  adequate  treatment  of  arches  may 
be  found  in  I.  O.  Baker's  "  Treatise  on  Masonry"  (John 
Wiley  &  Sons). 

Problem  186.  Find  the  shape  of  a  linkage  that  would  carry  the 
weights  of  Problem  178  with  the  same  horizontal  stress,  the  links 
being  in  compression.  Is  the  link  polygon  the  same  as  the  inverted 
string  polygon  of  that  problem  ? 

Problem  187.  Show  that  when,  and  only  when,  the  supports  are 
on  the  same  level,  the  link  polygon  for  a  given  set  of  loads  and 
horizontal  stress  is  the  inverted  string  polygon  for  that  set  of  loads. 
Show  also  that  in  any  case  the  link  polygon  can  be  obtained  from  the 
string  polygon  turned  through  180°  in  its  plane  if  in  constructing  the 
string  polygon  the  loads  and  their  horizontal  distances  are  reversed  in 
order  from  right  to  left. 

Problem  188.  An  arch  is  to  be  constructed  with  its  center  line 
in  the  form  of  an  arc  of  a  circle,  the  span  being  20  ft.,  and  rising  6  ft. 
in  the  middle.  Considering  the  load  on  the  arch  to  be  vertical  only 
and  at  any  point  proportional  to  the  distance  from  the  center  line  of 
the  arch  to  a  straight  horizontal  line  6  ft.  above  the  highest  point  of 
the  center  line,  sketch  approximately  the  line  of  resistance.  About 
how  deep  would  the  arch  stones  need  to  be  to  keep  the  line  of  resist- 
ance within  the  middle  third  ? 

SUGGESTION.  Assume  the  loads  concentrated  at  equal  horizontal 
intervals  and  construct  the  link  polygon  with  the  required  height  of 
6  ft.  It  may  then  be  seen  how  the  line  of  resistance  may  be  changed 


FLEXIBLE  CORDS  161 

so  as  to  more  nearly  coincide  with  the  given  circle  by  a  slight  move- 
ment of  the  pole. 

Problem  189.     Find  approximately  by  graphical  methods  the  form 
of  an  arch  to  carry  a  load  uniformly  distributed  along  the  horizontal. 

92.  Bending  Moments.  —  As  an  application  of  the  equi- 
librium polygon  for  parallel  forces  a  brief  discussion  is 
here  given  of  bending  moments  in  a  horizontal  beam  sub- 
jected to  vertical  forces  only.  Let  Fig.  146  represent  a 
horizontal  beam  acted  on 
by  vertical  forces.  At  any 
point  C  pass  a  vertical 
plane  perpendicular  to  the 
axis  of  the  beam. 

7? 

The  sum  of  the  moments  „ 

FIG.  146 

about  a  horizontal  line  in 

this  plane  of  all  the  forces  acting  on  the  beam  to  the  left 
of  the  plane  is  called  the  bending  moment  at  that  section. 
Clockwise  direction  will  be  counted  as  positive.  Repre- 
senting the  bending  moment  by  Jf,  the  value  of  M  at  0 
(Fig.  146)  is 


H — a,r*  «— a-»j 

T W///M 


Since  the  sum  of  the  moments  of  all  the  forces  acting  on 
the  beam  is  zero,  the  sum  of  the  moments  of  the  forces  to 
the  right  of  the  section  is  the  negative  of  the  sum  of  the 
moments  to  the  left  of  the  section.  Hence,  in  comput- 
ing bending  moments,  if  it  is  more  convenient,  we  may 
take  as  the  bending  moment  at  the  section  the  sum  of  the 
moments  of  all  forces  acting  on  the  beam  to  the  right  of 
the  section  and  consider  counter-clockwise  as  positive. 


162 


APPLIED  MECHANICS  FOR   ENGINEERS 


Problem  190.  A  beam  10  ft.  long  is  supported  at  the  ends  and 
carries  loads  of  500  lb.,  600  lb.,  800  Ib.  at  distances  of  3  ft.,  5  ft.,  and 
8  ft.  respectively  from  the  left  end.  Compute  the  bending  moment 
at  intervals  of  1  ft.  along  the  beam,  neglecting  the  weight  of  the 
beam. 

Problem  191.  A  beam  12  ft.  long,  supported  at  the  ends,  carries 
a  load  of  500  lb.  at  a  point  4  ft.  from  the  left  end  and  a  uniformly 
distributed  load  of  200  lb.  per  linear  foot  over  the  right  half  of  the 
beam.  The  beam  weighs  30  lb.  per  foot  of  length.  Compute  the 
bending  moment  at  intervals  of  2  ft.  along  the  beam.  Using  M  as 
ordinate  and  distance  along  the  beam  as  abscissa,  sketch  a  curve  repre- 
senting the  bending  moment  at  all  sections. 

93.  Bending  Moment  Diagram. —  Using  the  bending  mo- 
ment as  ordinate  and  distance  along  the  beam  as  abscissa, 
a  curve  may  be  plotted  which  shows  the  bending  moment 


FIG.  147 


at  each  section.  This  curve  is  called  the  bending  moment 
diagram.  It  will  now  be  shown  that  for  concentrated  loads 
the  equilibrium  polygon  represents  to  a  certain  scale  the 
bending  moment  diagram. 

Consider  a  beam  supported  at  the  ends  carrying  loads 
Pv  P2,  P3  lb.  at  distances  of  xv  #2,  xz  ft.  from  the  left  end 


FLEXIBLE  CORDS  163 

respectively.  Construct  the  equilibrium  polygon  for  the 
forces  and  reactions,  using  the  following  scale  : 

1"  =  m  ft.  along  the  beam, 
V  =  n  Ib.  on  the  line  of  loads. 

Let  D  be  the  pole  distance  in  inches,  and  d  the  depth  in 
inches  of  the  equilibrium  polygon  at  a  point  distant  —  in. 

Vfl 

from  the  left  support  between  the  loads  Pl  and  P2. 
By  similar  triangles 

m       n 

and  z  :  x~  x\  ==  — i :  ]). 

m  n 

Eliminating  z,  d  =  — —  1 — 1. 

By  definition  the  bending  moment,  M,  measured  in  pound- 
feet  at  the  given  point  is 

M=xEl-(x-xl)Pr 

.   d=    M  ^ 

or  M=mnD'd. 

This  formula  can  as  easily  be  shown  to  hold  for  any  other 
point.  Hence  the  depth  of  the  equilibrium  polygon  in  inches 
multiplied  by  the  number  of  ft.  per  inch  along  the  beam,  the 
number  of  Ib.  per  inch  on  the  load  line,  and  the  pole  distance 
in  inches  gives  the  value  of  the  bending  moment  in  Ib.-ft. 

The  depth  of  the  equilibrium  polygon  then  represents 
the  bending  moment  to  the  scale 

V  =  77w.ZHb.-ft. 


164 


APPLIED  MECHANICS  FOR   ENGINEERS 


If  it  is  desired  to  use  the  pound-inch  as  the  unit  of  moment, 
then  let  V  =m  in.  along  the  beam. 

To  obtain  the  bending  moment  diagram  to  a  given  scale, 
1"  =  k  lb.-ft.,  it  is  only  necessary  to  choose  D  so  that 
mnD  =  k. 

For  a  distributed  load  the  bending  moment  curve  may 
be  obtained  approximately  by  dividing  the  beam  length 
into  small  intervals  and  considering  the  load  on  each  in- 
terval to  act  at  the  center  of  that  interval. 

Problem  192.  Construct  graphically  the  bending  moment  for  the 
beam  of  Problem  190.  Let  1  in.  =  2  ft.  along  the  beam,  1  in.  =  500  Ib. 
on  the  load  line,  and  choose  D  so  that  1  in.  on  the  moment  diagram 
shall  equal  1500  lb.-ft.  of  moment. 

Problem  193.  Construct  approximately  the  bending  moment 
diagram  for  a  beam  supported  at  the  ends  and  carrying  a  uniformly 
distributed  load. 

94.  Bending  Moment  for  Beams  not  Supported  at  the  Ends. 
—  (a)  Cantilever  Beam.  A  cantilever  beam  is  shown  in 
Fig.  148.  The  construction  for  the  bending  moment  is 
indicated,  the  scale  factors  being  the  same  as  in  Art.  93. 


FIG.  148 


FLEXIBLE  COEDS 


165 


(5)  Overhanging  Beam.  In  Fig.  149  the  construction 
is  indicated  for  the  bending  moment  of  an  overhanging 
beam.  The  diagram  (i)  is  the  equilibrium  polygon. 
The  diagram  (ii)  is  obtained  from  (i)  by  replacing  the 
closing  line  of  the  polygon  and  the  two  lines  connected 


FIG.  149 

with  it  by  a  horizontal  line  and  drawing  a  polygon  with  a 
depth  on  each  load  line  and  reaction  line,  the  same  as  that 
of  the  equilibrium  polygon  (i). 

Problem  194.  Prove  that  the  method  indicated  of  drawing  the 
bending  moment  diagram  of  a  cantilever  beam  is  correct. 

Problem  195.  Prove  that  the  method  indicated  of  constructing 
the  bending  moment  diagram  of  the  overhanging  beam  is  correct. 

Problem  196.  Construct  the  bending  moment  diagram  of  a 
cantilever  beam  12  ft.  long  carrying  loads  of  400,  600,  and  1000  Ib. 
at  distances  of  0,  3,  and  7  ft.  respectively  from  the  free  end.  Scale 
the  bending  moment  at  3  ft.  and  at  6  ft.  from  the  free  end  and  com- 
pare with  computed  values.  State  the  scales  used. 


166  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  197.  A  beam  16  ft.  long  is  supported  at  two  points  dis- 
tant respectively  3  ft.  from  the  left  end  and  4  ft.  from  the  right  end. 
Loads  of  300,  500,  1000,  and  600  Ib.  are  applied  at  distances  of  0,  6, 
10,  and  16  ft.  respectively  from  the  left  end.  Construct  graphically 
the  bending  moment  diagram  to  a  convenient  scale.  Check  by  com- 
puting analytically  the  bending  moment  at  several  points. 

95.  Tensile  and  Compressive  Stresses  in  Beams.  —  Let  Fig. 
150  represent  the  portion  of  the  beam  of  Fig.  146  to  the 
left  of  the  plane  section  at  (7.  This  portion  of  the  beam 

is  in  equilibrium  under  the  action 
of  the  external  forces  acting  on 
this  portion  and  the  internal  forces 
—    exerted  on  it  by  the  portion  of  the 
beam  to  the  right  of  the  section. 
These  internal  forces  can  be  re- 
F     150  solved  into  horizontal  and  vertical 

forces.     The  vertical  component  of 

the  internal  forces  must  be  equal  and  opposite  to  the  sum 
of  the  external  forces  to  the  left  of  the  section.  This 
latter  sum  is  called  the  shear  at  the  section. 

Since  the  external  forces  are  all  vertical,  the  sum  of  the 
horizontal  components  of  the  internal  forces  must  be  zero. 
The  horizontal  forces  therefore  form  a  couple.  This 
couple  is  called  the  internal  stress  couple.  To  produce 
equilibrium  the  moment  of  the  internal  stress  couple  must 
be  equal  and  opposite  in  sign  to  the  moment  about  any 
horizontal  line  in  the  section  of  all  the  external  forces  to 
the  left  of  the  section.  That  is,  the  internal  stress  couple 
at  any  section  of  the  beam  is  equal  to  the  bending  moment 
at  that  section  changed  in  sign. 


FLEXIBLE   CORDS 


167 


It  can  be  shown  by  experiment  that  when  a  rod  is  stretched 
or  compressed  within  the  elastic  limit,  the  amount  of 
elongation  or  compression  is  directly  proportional  to  the 
applied  force.  Denning  unit  stress  in  a  rod  uniformly 
stretched  or  compressed,  i.e.  without  bending,  as  the  total 
force  divided  by  the  cross-sectional  area,  and  unit  strain 
as  the  amount  of  elongation  or  compression  divided  by  the 
original  length,  this  experimental  fact  may  be  expressed 

by  the  equation, 

unit  stress       -^ 

• =  Jjj 

unit  strain 

where  J^is  constant  for  a  given  material. 

E  is  called  the  modulus  of  elasticity.  Experiment  shows 
that  it  is  the  same  for  compression  as  for  tension,  for  such 
materials  as  wood  and  structural  steel. 

It  is  also  shown  by  experiment  that  when  a  beam  is 
bent,  within  the  elastic  limit,  a  plane  section  of  the  beam 
before  bending  remains  a  plane  section  after  bending. 
When  the  beam  is  bent,  the  upper  fibers  of  the  beam  are 
then  compressed  (or  elongated)  and  the  lower  fibers  are 
elongated  (or  compressed)  as  in  Fig.  151.  There  will  be 


/        B' 


A          £ 


A  B 


FIG.  151 


168  APPLIED  MECHANICS  FOR  ENGINEERS 

somewhere  between  top  and  bottom  a  surface,  called  the 
neutral  surface,  where  there  is  neither  tension  nor  com- 
pression, and  the  amount  of  tensile  or  compressive  stress  at 
any  point  in  the  section  will  vary  directly  as  the  distance 
of  this  point  from  the  neutral  surface,  since  the  amount  of 
the  elongation  or  compression  of  the  longitudinal  fiber  of 
the  beam  through  the  point  varies  as  the  distance  from 
the  neutral  surface. 

If  s  is  the  unit  compressive  stress  at  the  upper  outside 
fiber  of  the  beam,  distant  y  from  the  neutral  surface,  s'  the 
unit  stress  at  a  distance  y1  from  the  neutral  surface,  then 


sr  will  be  positive  or  negative  according  to  the  sign  of  yf  . 
Considering  the  portion  of  the.  beam  as  in  equilibrium 
under  the  action  of  the  external  and  internal  forces,  if  dA 
is  an  element  of  area  of  the  cross  section  distant  y'  from 
the  neutral  surface,  we  have  from  2JT=  0, 


or 

But  \y'dA  is  equal  to  the  product  of  the  area  of  the  cross 
section  and  the  ordinate  of  the  center  of  gravity  measured 
from  the  origin  of  ordinates,  i.e.  from  the  neutral  surface. 
If  y  is  this  ordinate,  then 


Therefore 


FLEXIBLE  CORDS  169 

and  the  neutral  surface  passes  through  the  center  of 
gravity  of  the  section.* 

Equating  moments  of  external  and  internal  forces  about 
a  horizontal  gravity  axis  of  the  section, 

Mom  of  external  forces  =  —  mom  of  internal  forces, 


'dA 


or 


jr  y 

where  /  is  the  moment  of  inertia  of  the  area  of  the  cross 
section  with  respect  to  the  horizontal  gravity  axis  of  the 
section. 


This  formula  gives  the  stress  s  at  a  distance  y  from  the 
neutral  axis  of  the  section.  The  stress  at  any  other  point 
in  the  section  is  obtained  by  the  direct  proportion 

!U£ 

s       y' 

For  a  beam  of  constant  cross  section  y  and  /are  constant 
and  hence  8  varies  directly  as  M.  The  maximum  tensile 
or  compressive  stress  will  therefore  be  found  where  the 
bending  moment  has  maximum  numerical  values. 

In  computing  stresses  care  must  be  used  to  have  the 
same  units  throughout  the  formula.  If  stress  is  reckoned 
in  pounds  per  square  inch,  the  dimensions  of  the  cross 

*  In  the  above  it  is  assumed  that  the  intersection  of  the  vertical  section 
and  the  neutral  surface  is  a  horizontal  straight  line.  If  the  loads  lie  in 
one  plane  which  is  a  plane  of  symmetry  of  the  beam,  this  will  be  the  case. 
Under  other  conditions  the  neutral  surface  may  not  be  horizontal. 


170  APPLIED  MECHANICS  FOR   ENGINEERS 

section  must  be  in  inches  and  the  bending  moment  in 
pounds-inches. 

Problem  198.  A  beam  4  in.  broad  by  6  in.  deep,  12  ft.  long,  is 
supported  at  the  ends  and  carries  loads  of  400  Ib.  and  800  Ib.  at  dis- 
tances of  5  ft.  and  9  ft.  respectively  from  the  left  end.  Sketch  the 
bending  moment  diagram  and  find  the  maximum  fiber  stress. 

Ans.     1150  Ib.-sq.  in.  under  the  800  Ib.  load. 

Problem  199.  A  beam  4  in.  broad  by  6  in.  deep,  12  ft.  long,  is 
supported  at  the  ends  and  carries  two  equal  loads  at  distances  of  4  ft. 
from  the  ends.  What  maximum  value  may  the  loads  have  if  the 
maximum  allowable  fiber  stress  is  1000  Ib.  per  sq.  in.  ? 

Problem  200.  Find  the  ratio  of  the  loads  that  could  be  put  on  a 
beam  4"  by  6"  when  placed  with  the  4"  face  horizontal  and  when 
placed  with  the  6"  face  horizontal,  the  maximum  fiber  stress  to  be 
the  same  in  the  two  cases. 

Problem  201.  Compare  the  loads,  one  concentrated  at  the  middle, 
the  other  uniformly  distributed  over  the  length  of  the  beam,  to  cause 
the  same  maximum  fiber  stress  in  the  beam. 

Problem  202.  An  I-beam  of  depth  10  in.  and  weight  25  Ib.  per 
linear  foot  is  16  ft.  long  and  is  supported  at  the  ends.  What  con- 
centrated load  can  it  carry  in  the  middle  so  as  to  cause  a  maximum 
fiber  stress  of  16,000  Ib.  per  sq.  in.?  The  moment  of  inertia  of  the 
cross  section  about  a  horizontal  gravity  axis  is  122  in.4.  Take  the 
weight  of  the  beam  into  consideration. 

96.  Cords  and  Pulleys.  —  When  a  cord  passes  over  a 
pulley,  without  friction,  the  tension  is  transmitted  along 
its  length  undiminished.  A  weight  W  attached  to  a  cord 
which  passes  vertically  over"  a  pulley  is  raised  by  a  direct 
downward  pull  P  on  the  other  end  of  the  rope.  If  there 
is  no  friction,  P  is  equal  to  IF,  for  uniform  motion.  In 
the  case  of  a  system  of  pulleys,  as  shown  in  Fig.  152,  the 


FLEXIBLE  COEDS 


171 


B 


cord  may  be  considered  as  under  the  same  tension  through- 
out and  parallel  to  itself  in  passing  from  one  sheave  to  the 
other.  It  is  then  possible  to  cut  across 
the  cords,  just  as  was  done  in  the  case  of 
the  bridge  truss,  Problem  92,  where  the 
stress  was  along  the  member  in  each  case. 
Cutting  all  the  cords  at  0  and  considering 
all  the  forces  acting  on  the  sheave  B,  we 
get,  calling  the  tension  in  the  cord  P, 

6  P  =  W, 

or  the  tension  in  the  cord  is  W/6.  A  con- 
sideration of  the  upper  sheaves  gives 
T=  1  P  =  7/6  (  W).  The  various  cases 
of  cords  and  pulleys  that  come  up  in 
engineering  work  may  be  taken  up  in  a 
similar  way,  but  in  any  case  of  cutting 
cords,  it  must  be  remembered  that  all 
cords  attaching  one  part  to  another  must 
be  cut  and  the  tension  acting  along  the  cords  inserted  be- 
fore the  principles  of  equilibrium  can  be  applied.  The 
consideration  of  the  friction  between  cords  and  pulleys  will 
be  taken  up  in  Chapter  XIII. 

Problem  203.  Suppose  the  hook  of  the  lower  sheave  of  Fig.  152 
attached  to  a  weight  of  350  lb.,  and  a  man  of  weight  150  Ib.  stands 
on  the  weight  and  lifts  himself  and  the  weight  by  pulling  on  the  rope. 
What  force  must  he  exert?  If  he  stands  oil  the  ground,  what  force 
must  he  exert  to  raise  a  weight  of  500  lb.  ? 

97,  Equilibrium  of  a  Flexible  Cord  Carrying  a  Load  Uniformly 
Distributed  along  the  Horizontal.  —  Let  the  cord  supported  at 


W 


FIG.  152 


172 


APPLIED  MECHANICS  FOR  ENGINEERS 


A  and  B  (Fig.  153  (a)  and  (6))  carry  a  uniform  load  w  per 
unit  length  along  the  horizontal.  Consider  a  small  portion 
of  the  cord,  As,  with  ends  at  (x,  y}  and  (x  +  A#,  y  +  A?/), 
the  coordinate  axes  being  horizontal  and  vertical.  This 


(a) 


portion  is  in  equilibrium  under  the  action  of  the  tensions  T 
and  T'  at  its  ends  and  the  load  w&x  carried  by  this  portion 
of  the  cord  (Fig.  153  (<?)). 

Resolving  T  and  T'  into  horizontal  and  vertical  compo- 
nents and  applying  the.  conditions  for  equilibrium,  we  get 


But 


Y'  -  r=  wkx. 

Yr  —  Y=  increment  of  F=AF. 


(1) 
(2) 


Equation  (1)  shows  that  the  horizontal  component  of  the 
tension  is  the  same  throughout  the  cord,  and  equation  (2) 
may  be  written 

dY 

T~  =  w> 
dx 

and  hence,  Y=  wx  +  c.  (3) 

Now  X  and  Y  are  respectively  equal  to  Tcos  6  and  T 
sin  0,  where  6  is  the  inclination  of  the  tangent  line  to  the 


FLEXIBLE  CORDS  173 

curve   at   (x,  y).     Equations   (1)   and  (3)  may  then  be 

written 

/    Tcos  6  =  X, 

Tsiu  0  =  wx  +  c. 


By  division  we  get,  since  tan  0  =  - 

d 


dy     wx  +  c 
dx          X 

Since  c  and  w  are  constants,  a  value  of  x  can  be  found  for 
a  point  on  the  curve,  not  necessarily  on  the  cord  between 

A  and  B,  for  which  -^  =  0.     If  the  origin  be  moved  to  this 
dx 

point  on  the  curve  by  a  translation  of  axes,  equation  (4) 
will  take  the  form 


This  means  that  the  origin  is  taken  at  the  lowest  point  of 
the  curve  as  in  Fig.  153  (a)  and  (6).     In  (a)  the  origin 
is  on  the  cord,  and  in  (6)  it  is  not. 
Integrating  equation  (5), 


But  y  —  0  when  x  =  0.     Therefore  c'  =  0  and  the  equation 
of  the  curve  becomes 


Hence  the  cord  hangs  in  a  portion  of  a  parabola  with  ver- 
tical axis. 

98.   The  Horizontal  Component  of  the  Tension  in  Terms  of 
the  Length  of  Span  and  the  Deflection. 

A.    E^^^J^^"1^  *»        """  J 

n  7 


A 


174  APPLIED  MECHANICS  FOB   ENGINEERS 

The  horizontal  distance  between  the  supports  is  called 
the  span,  I. 

The  vertical  distance  of  the  vertex,  of  the  parabola 
below  the  lower  point  of  support  is  called  the  deflection,  d. 

In  Fig.  153  (a)  and  (£)  the  coordinates  of  A  are 
(  —  l^  d)  and  (lv  d)  respectively,  and  the  coordinates  of 
E  are  (72,  d  +•  K).  The  coordinates  of  A  and  B  satisfy 
the  equation  of  the  curve.  Hence  in  case  (a) 


Therefore,      ^ 

V 

and  J*T=^ 


2(yS  +  h  +  Vd)2 

If  h  =  0,  the  supports  are  on  the  same  level  and 

X=%d' 

In  case  (ft),  1  =  12  —  ll== 
and  JT  = 

99.   Length  of  Cord. — The  length  of  the  curve  from  the 
origin  to  any  point  (x,  y)  on  the  curve  is  given  by 


FLEXIBLE  CORDS  175 


Letting     —  =  a,    the    equation  of  the  parabola   becomes 
w 

y  =  ^—  ,  from  which  -^  =  -,  and  hence 
2  a  dx      a 


+  a2  log/*  4-  Va2  + 


The  total  length  of  the  curve  is 
in  case  (a)     S  —  s±  +  s2,  and 


in  case  (o) 

Another   formula    for   the   length   of   the    cord   may  be 

obtained  by  expanding  "vl +— by  the  binomial    theorem 

a* 

and  integrating  the  series  term  by  term  when  a  is  greater 
than  each  of  the  quantities  ^  and  12* 

i 

Thus,        «!=  P(l  +  — )<fo. 
Jo  V        #  / 

1  #      1  x±       1  3*        5    a* 

2  a2      8  a4      16a6      128  a8 


. 
6  a2      40  a4      112  a6      1152  a8 

In  terms  of  the  deflection,  c?,  this  becomes,  since  a  =  — *-, 


176  APPLIED  MECHANICS  FOB  ENGINEERS 

In  like  manner,  if  d±  =  d  +  A, 


If  the  supports  are  on  the  same  level,  Zj  =  lz  =  -  and  the 
expressions  for  the  total  length  S  become  respectively 

*_ir'v*rF4P+*iog.'-±4±ir 

a[_4:  2a 


, 

and 


3  23 .  a2     20  25  .  a4     56 

93  .  ,72        05  .  /74        9  .  97  . 


. 


For  cords  for  which  the  deflection  is  small  compared  to 
the  length  of  span  the  series  converge  rapidly  and  three 
or  four  terms  are  sufficient  for  computing  the  length. 
For  large  values  of  d  the  series  should  not  be  used. 

Problem  204.  A  suspension  bridge  as  shown  in  Fig.  154  has  a 
span  of  1200  ft.  and  uhe  cable  a  maximum  deflection  at  the  center 
d  =  120  ft.  The  weight  of  the  floor  is  2  tons  per  linear  foot.  Find 


FIG.  154 


FLEXIBLE  COEDS  177 

the  equation  of  the  cable  and  the  tension  at  0  and  at  B.  If  the  safe 
strength  of  cable  is  75,000  Ib.  per  sq.  in.,  find  the  area  of  wire  section 
of  cable  necessary  to  support  the  floor. 

Problem  205.  Find  the  length  of  the  cable  in  the  preceding 
problem. 

Problem  206.  A  cable  is  to  be  suspended  from  two  points  distant 
100  ft.  apart  horizontally  and  20  ft.  vertically.  It  is  to  carry  a  load  of 
500  Ib.  per  horizontal  foot,  and  the  lowest  point  of  the  curve  is  to  be 
25  ft.  below  the  lower  point  of  support.  Find  the  position  of  the 
lowest  point  of  the  cord,  the  value  of  the  horizontal  component  of 
the  tension,  and  the  maximum  tension. 

Problem  207.  Find  the  length  of  the  cable  of  the  preceding 
problem. 

Problem  208.  A  cable  105  ft.  long,  carrying  a  uniform  horizontal 
load,  is  stretched  between  two  points  on  the  same  level  distant  100  ft. 
apart.  Find  the  deflection. 

SUGGESTION.  Use  the  first  three  terms  of  the  series  which  expresses 
S'in  terms  of  I  and  d.  See  how  large  the  fourth  term  is  with  the 
computed  value  of  d. 

Problem  209.  A  uniform  wire  weighing  \  Ib.  per  foot  of  length  is 
supported  between  two  points  200  ft.  apart  on  the  same  level  and  the 
maximum  deflection  is  2|  ft.  Find  the  horizontal  tension,  the 
maximum  tension,  and  the  length  of  the  wire. 

SUGGESTION.  Since  the  deflection  is  here  relatively  small,  the  load 
is  very  nearly  uniformly  distributed  along  the  horizontal.  The  curve 
may  then  be  regarded  as  approximately  a  parabola. 

100.  Equilibrium  of  a  Flexible  Cord  Carrying  a  Load  Uni- 
formly Distributed  along  the  Cord.  —  Using  a  notation  like 
that  of  Art.  97,  the  same  method  leads  to  an  equation 
like  (5)  of  that  article,  with  x  replaced  by  s;  namely, 

dy  _w^ 
dx      X' 


178  APPLIED   MECHANICS  FOR  ENGINEERS 

in  which  the  origin  is  at  a  point  of  the  curve  where  the 
slope  is  zero  and  s  is  the  length  of  the  curve  from  the 
origin  to  the  point  (#,  y). 

Y 


(a) 


00 

FIG.  155 


Replacing  dy  by   Vds2  —  dx*  and   writing  —  =  #,  this 

w 

becomes 


or 


ds  =  -  Va2  H- 
a 

ds  dx 


•  dx. 


+  s2       *    ' 


Integrating,         loge 

a 

s  =  0  when  x  =  0  ;   .  •.  c  =  log  a. 
log. 


Invert, 


_£ 
=  e~*. 


*  +  Va2  +  s2 


FLEXIBLE  CORDS  179 

Rationalizing  the  denominator, 


A11. 
Add  to 


a 

X  X 


Then  s==|(c«_e  «).  (1) 

It  was  found  above  that  -^  =  -. 

&      a 


CL       —  —  — 

from  which  y  =  -  (e«  +  g  «)+<?'. 


But  a;  =  0wheny  =  0;     .•.  c'  =  —  a. 


which  is  therefore  the  equation  of  the  curve,  the  origin 
being  at  the  lowest  point  of  the  curve  (Fig.  155  (a)  and 

(*»• 

This  curve  is  called  the  catenary. 

The  length  of  the  curve  follows  at  once  from  equation 
(1)  ;  namely, 

04-  «i»|(««--«""«) 

and  OB  ««,»|(*«-«*),  (Fig.  155) 

and  S=  s2±  81  according  as  the  cord  hangs  as  in  (a)  or  (6) 
of  Fig.  155. 


180  APPLIED  MECHANICS  FOR  ENGINEERS 


If  the  points  of  support  are  on  the  same  level,  \  =  ?2  ==  ^, 
and  the  total  length  of  cord  is 


The  values  of  y  and  s  may  be  expressed  in  a  series  by 
making  use  of  the  expansion 

•    '-'+•+1+1+1+1+-  •/: 

It  is  left  for  the  student  to  show  that 


- 

|_3_  a2      |5  a4 

101.  Eepresentation  by  Means  of  Hyperbolic  Functions.  — 
From  the  definitions 

sinh  x  =  \  (ex  —  e~x), 
cosh  x=\  (ex  +  e~x), 

it  follows  at  once  that  the  values  of  y  and  *,  the  arc  from  0 
to  the  point  (#,  #),  are  given  by  the  formulae 

y  -h  a  =  a  cosh-, 
a 

8  =  a  sinh—,        where  a  =  —  • 
a  w 

A  table  of  hyperbolic  sines  and  cosines  is  given  in  the  ap- 
pendix and  should  be  used  in  the  solution  of  the  following 
problems. 

Problem  210.  A  flexible  wire  weighing  \  Ib.  per  foot  is  supported 
by  two  posts  200  ft.  apart.  The  horizontal  pull  on  the  wire  is  500  Ib. 
Find  the  deflection  at  the  center  and  the  length  of  the  wire. 


FLEXIBLE  CORDS  181 

Problem  '211.  What  pull  will  be  necessary  in  Problem  210  so  that 
the  greatest  deflection  will  not  be  greater  than  6  in.  ?  What  is  the 
length  of  the  wire  for  this  case? 

Problem  212.  Find  the  tension  in  the  wire  of  Problem  210  at  the 
supports.  -T~  M>U  l  -|: 

Problem  213.  A  wire  weighing  I  Ib.  per  foot  is  suspended  be- 
tween two  points  A  and  B  where  B  is  20  ft.  higher  than  A  and  distant 
120  ft.  horizontally.  The  horizontal  component  of  the  tension  of  the 
wire  is  50  Ib.  Find  the  position  of  the  lowest  point  of  the  curve,  the 
deflection,  the  length  of  the  wire,  and  the  maximum  tension.  Sketch 
the  curve  in  which  the  wire  hangs. 

SUGGESTION.  Assuming  that  the  wire  hangs  as  in  (a),  Fig.  155, 
we  may  write 

200  +  d  +  20  =  200  cosh  A ,      since  a  =  -  =  200, 
4\)0  w 


200 +rf  =  200  cosh  ^0. 
Subtracting,      20  =  200  f  cosh  ^  -  cosh  ^ 


since  cosh  x  —  cosh  y  =  2  sinh  x  "*"  y  sinh  - 

2  2 

But  Zj  +  12  =  120. 

Therefore  20  =  400  sinh  .3  .  sinh  l- 


from  which  the  value  of  12  —  Zx  may  be  found  from  the  table  of  hyper- 
bolic sines.  Combining  this  value  with  12  +  Zj  =  120,  the  values  of  £, 
and  /2  are  determined.  The  values  of  d,  S,  and  the  maximum  tension 
may  then  be  determined. 

Problem  214.     Solve  the  preceding  problem  if  B  is  40  ft.  higher 
than  A,  all  other  conditions  remaining  unchanged. 


182  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  215.     A  wire  is  suspended  between  two  points  on  the 

same  level  240  ft.  apart.     The  deflection  at  the  center  is  60  ft.     Find 

j^ 
the  value  of  a,  or  — ,  on  the  supposition  that  the  load  is  (1)  uniformly 

w 

distributed  along  the  horizontal,  (2)  uniformly  distributed  along  the 
wire.  Compute  the  values  of  y  for  x  =  40  and  for  x  =  80  ft.  for  both 
cases  and  sketch  the  parabola  and  the  catenary. 

SUGGESTION.  In  case  (1)  a  is  obtained  by  substituting  the  coordi- 
nates of  the  point  of  support  in  the  equation  of  the  parabola.  In  case 
(2)  the  value  of  a  found  for  the  parabola  may  be  used  as  a  trial  value 
and  the  correct  value  of  a  found  by  modifying  the  trial  value  until  a 
value  is  found  to  satisfy  the  equation 

y  +  a  =  a  cosh  - ,        where  y  =  60  and  x  =  120. 


That  is,  a  must  satisfy 


59+1=  cosh  I??. 


Problem  216.    Solve  the  preceding  problem  if  the  deflection  in  the 
middle  is  only  6  ft. 


<T-     S  P 

y    -•  6-& 

CHAPTER    IX 
MOTION  IN  A  STRAIGHT  LINE    (RECTILINEAR  MOTION) 

102.  Velocity.  —  When  a  particle  moves  along  a  straight 
line  passing  over  equal  spaces  in  equal  times,  it  is  said  to 
have  uniform  motion.  In  this  case  the  ratio  of  the  space 
passed  over  to  the  time  taken  to  pass  over  that  space  is 
called  the  velocity  of  the  particle. 

If  the  motion  is  along  a  straight  line  but  is  not  uniform, 
the  ratio  of  the  space  passed  over  in  any  time  to  the  time 
is  called  the  average  velocity  of  the  particle  for  that  time, 
or  space.  Thus,  if  the  particle  moves  from  P  to  Pv  a 

distance  As,  in  the  time  A£,  then  —  =  average  velocity  of 
the  particle  between  P  and  Pv 

The  limiting  value  of  —  as  A*  approaches  the  limit  zero 
is  defined  to  be  the  velocity  of  the  particle  at  P. 
Thus,  v=!f; 

that  is,  the  velocity  is  the  first  derivative  of  the  distance  with 
respect  to  time. 

The  unitt)f  velocity  is  the  unit  of  space  per  unit  of  time, 
as  ft.  per  sec.,  mi.  per  hr.,  etc.  Speed  is  sometimes  used 
instead  of  velocity,  especially  in  speaking  of  the  motion  of 
machines  or  parts  of  machines.  Speed,  however,  involves 
only  the  rate  of  motion  without  reference  to  its  direction, 
while  velocity  involves  both  rate  of  motion  and  the  direc- 
tion in  which  the  motion  takes  place. 

183 


184  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  217.  The  space  passed  over  by  a  particle  moving  in  a 
straight  line  is  given  by  the  formula  s  —  16  t2,  where  s  is  the  distance 
in  feet  passed  over  in  the  time  t  in  seconds.  Find  the  velocity  when 
t  =  0,  when  t  =  2,  and  at  the  end  of  t  sec.  What  is  the  velocity  of  the 
particle  when  it  has  moved  40  ft.  ? 

Problem  218.  A  particle  moves  back  and  forth  along  a  straight 
line,  the  abscissa  of  the  particle  being  given  by  x  =  k  cos  (ct).  Find 

Q 

the  location  of  the  particle  and  its  velocity  when  t  =  0.  —  ,  -,  —  . 

2c    c    2c 

Erect  ordinates  to  represent  the  velocity  for  several  positions  of  the 
moving  particle  and  represent  the  velocity  by  a  curve. 

103.  Acceleration  for  Straight-line  Motion.  —  When  the 
velocity  of  a  particle  moving  in  a  straight  line  increases 
by  equal  amounts  in  equal  times,  the  motion  is  said  to  be 
uniformly  accelerated,  and  the  gain  in  velocity  per  unit  of 
time  is  called  the  acceleration  of  the  particle. 

If,  for  straight-line  motion,  the  velocity  changes  from  v 

at  the  time  t  to  v  +  At>  at  the  time  t  +  A£,  then  —  is  called 

A£ 

the  average  acceleration  of  the  particle  for  the  time  A£. 

The  limiting  value  of  —  ,  i.e.   —  ,  is  defined  to  be  the 
A£  dt 

acceleration  of  the  particle  at  the  time  t. 
Thus,  if  a  represents  acceleration, 


=:       = 
dt      dt2 

Another  form  for  a  is  frequently  used.     Since  a  =  —  -, 

ctt 

and  v  =  —  ,  the  value  of  a  may  be  written 
dt 

_dv  _dv  ds 

~  "  d£~  'fodt' 

.-„* 


MOTION  IN  A   STRAIGHT  LINE  185 

The  unit  of  acceleration  is  the  unit  of  velocity  per  unit 
of  time,  as  ft.  per  sec.  per  sec.,  mi.  per  hr.  per  hr.,  yd. 
per  min.  per  sec.,  etc. 

Problem  219.  Find  the  acceleration  of  the  particle  of  Problem 
217  at  the  times  specified.  Derive  an  equation  which  expresses  the 
velocity  in  terms  of  the  space,  and  from  this  equation  obtain  a 
formula  for  the  acceleration  in  terms  of  the  space. 

Problem  220.  Using  t  for  abscissa  and  s,  y,  a,  respectively,  as  or- 
dinates,  plot  the  curves  which  represent  the  space,  velocity,  and  ac- 
celeration respectively  in  terms  of  the  time  in  Problem  217.  Show  how 
the  ordinate  to  the  velocity  curve  is  related  to  the  slope,  or  gradient,  of 
the  space  curve  at  any  value  of  t.  Likewise  for  the  acceleration  and 
velocity  curves. 

Problem  221.  Find  the  acceleration  of  the  moving  particle  in 
Problem  218.  Show  that  the  acceleration  is  proportional  to  x.  Plot 
curves  showing  x,  v,  and  a  in  terms  of  t.  Also  a  curve  showing  a  in 
terms  of  x. 

Problem  222.  Show  that  if  a  curve  is  plotted,  using  values  of 
time  as  abscissas  and  values  of  velocity  as  ordinates,  the  area  under 
the  curve  between  two  values  of  the  time  is  equal  to  the  space  passed 
over  by  the  particle  in  that  interval  of  time. 

104.  Constant  Acceleration.  —  When  the  acceleration  is 
constant,  we  have  the  relation  dv  =  acdt,  ac  representing 
the  constant  value  of  a,  and  therefore 


I  vdv  =  ac  I  dt, 

«/P0  c/O 


i    .  ds 

and  since  v  =  — 

dt 


or 


186  APPLIED  MECHANICS  FOR  ENGINEERS 

In  a  similar  way  the  relation 

gives  ds 

(  vdv  =  ac  J  "ds ; 

»X  *  A  »XO 

therefore  ^ —  -^  =  «c«, 

o  o 

V2  —  V2 

These  equations  of  motion  give  the  velocity  in  terms  of 
time,  the  distance  in  terms  of  time,  and  the  distance  in 
terms  of  velocity. 

105.  Freely  Falling  Bodies.  —  Bodies  falling  toward  the 
earth  near  its  surface  have  a  constant  acceleration.  It  is 
usually  represented  by  g  and  equals  approximately  32.2  ft. 
per  second  per  second.  The  value  of  g  varies  slightly  with 
the  height  above  the  sea  level  and  the  latitude,  but  for 
the  purposes  of  engineering  it  may  usually  be  taken  as 
32.2.  The  equations  of  motion  for  such  bodies  are,  then, 


If  the  body  falls  from  rest,  00  =  0,  and  the  equations  of 
motion  become 


This  latter  is  often  written  v2  =  2A,  where  h  =  8. 


MOTION  IN  A   STRAIGHT  LINE  187 

106.  Body  Projected  Vertically  Upward.  —  When  a  body 
is  projected  vertically  upward  from  the  earth,  the  accelera- 
tion is  constant  and  equals  —  g.  If  the  velocity  of  pro- 
jection is  #0,  the  equations  of  motion  are 


Problem  223.  A  body  is  projected  vertically  downward  with  an 
initial  velocity  of  30  ft.  per  second  from  a  height  of  100  ft.  Find 
the  time  of  descent  and  the  velocity  with  which  it  strikes  the  ground. 

Problem  224.  A  body  falls  from  rest  and  reaches  the  ground  in 
6  sec.  From  what  height  does  it  fall,  and  with  what  velocity  does 
it  strike  the  ground  ? 

Problem  225.  A  body  is  projected  vertically  upward  and  rises  to 
the  height  of  200  ft.  Find  the  velocity  of  projection  v0  and  the  time 
of  ascent.  Also  find  the  time  of  descent  and  the  velocity  with  which 
the  body  strikes  the  ground. 

Problem  226.  A  stone  is  dropped  into  a  well,  and  after  2  sec. 
the  sound  of  the  splash  is  heard.  Find  the  distance  to  the  surface  of 
the  water,  the  velocity  of  sound  being  1127  ft.  per  second. 

Problem  227.  A  man  descending  in  an  elevator  whose  velocity 
is  10  ft.  per  second  drops  a  ball  from  a  height  above  the  elevator 
floor  of  6  ft.  How  far  will  the  elevator  descend  before  the  ball 
strikes  the  floor  of  the  elevator  ? 

Problem  228.  In  the  preceding  problem,  suppose  the  elevator 
going  up  with  the  same  velocity,  find  the  distance  the  elevator  goes 
before  the  ball  strikes  the  floor  of  the  elevator. 

107.  Newton's  Laws  of  Motion.  —  Three  fundamental  laws 
may  be  laid  down  which  embody  all  the  principles  in 
accordance  with  which  motion  takes  place.  These  are 


188  APPLIED  MECHANICS  FOR   ENGINEERS 

the  result  of  observation  and  experiment  and  are  known 
as  Newton's  Laws  of  Motion. 

First  Law.  Every  body  remains  in  a  state  of  rest  or 
of  uniform  motion  in  a  straight  line  unless  acted  upon  by 
some  unbalanced  force. 

Second  Law.  When  a  body  is  acted  upon  by  an  unbal- 
anced force,  the  acceleration  is  along  the  line  of  action  of 
the  force,  and  is  proportional  to  the  force  applied  and  in- 
versely proportional  to  the  mass  of  the  body. 

Third  Law.  To  every  action  of  a  force  there  is  always 
an  equal  and  opposite  reaction. 

The  second  law  states  that  in  case  the  system  of  forces 
acting  on  the  body  is  unbalanced,  the  motion  is  accelerated. 
The  acceleration  is  in  the  direction  of  the  resultant  force 
and  is  proportional  to  it.  It  also  implies  that  each  force 
of  the  system  produces  or  tends  to  produce  an  acceleration 
in  its  own  direction  proportional  to  the  force.  That  is  to 
say,  each  force  produces  its  own  effect,  regardless  of  the 
action  of  the  other  forces. 

As  a  result  of  this  latter  fact,  if  a  body  is  acted  upon 
by  a  force  P  and  the  earth's  attraction  #,  we  have 
P:  a  =  a:g, 

where  6r  is  the  weight  of  the  body,  g  the  acceleration  of  grav- 
ity, and  a  the  acceleration  due  to  the  force  P.     From  this  it 

follows  that  G 

P  =  —  *a  =  M-a. 

ff 
/nr 

The  quantity  — ,  in  which  G-  is  the  weight  of  the  body 

9 

in  pounds,  and  g  is  in  ft. /sec.2,  is  called  the  mass  of  the 
body  in  "  Engineer's  Units." 


MOTION  IN  A   STRAIGHT  LINE  189 

If  ^  =  32.2,  the  Engineer's  Unit  of  mass  is  equivalent 
to  32.2  pounds. 

108.    Motion    on    an    Inclined    Plane.  —  A     body    (See 
Fig.  156),  of  weight  6r,  moves  down  an  inclined  plane, 
without    friction,   under    the    action    of 
a  force  6r  sin  6.     The  acceleration  down 
the      plane      equals      the      accelerating 
force  divided   by  the  mass   (Art.    107) 

=  —  -~  —  —g  sin  0.       The  acceleration  is 

vr  FIG.  156 

9 

constant.      The  equations  of  motion  for  such  a  case,  then, 

are  (Art.  104) 

v  =  (g  sin  6)  t  +  v0, 


If  the  body  starts  from  rest  down  the  plane,  VQ  =  0.  If 
it  be  projected  up  the  plane  with  an  initial  velocity  v0,  the 
acceleration  equals  —  g  sin  6. 

Problem  229.  A  body  is  projected  up  an  inclined  plane  which 
makes  an  angle  of  60°  with  the  horizontal  with  an  initial  velocity  of  12 
ft.  per  second.  -Neglecting  friction  of  the  plane,  how  far  up  the  plane 
will  the  body  go  ?  Find  the  time  of  going  up  and  of  coming  down. 

Problem  230.  A  body,  weight  20  lb.,  is  projected  down  the  plane 
given  in  the  preceding  problem  with  a  velocity  of  20  ft.  per  second. 
How  far  will  it  go  during  the  third  second?  ^  '  \ 

Problem  231.  Suppose  the  'body  in  the  preceding  problem 
meets  a  constant  force  of  friction  F  =  10  lb.  What  will  be  the 
acceleration  down  the  plane  ?  How  far  will  it  go  during  the  second 
second  ? 


J  I. 


190 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  232.  A  boy  who  has  coasted  down  hill  on  a  sled  has  a 
velocity  of  10  mi.  per  hour  when  he  reaches  the  foot  of  the  hill. 
He  now  goes  on  a  horizontal,  meeting  a  constant  resistance  of  25  Ib. 
If  the  combined  weight  of  the  boy  and  sled  is  75  Ib.,  how  far  will  he 
go  before  coming  to  rest?  I  0*  0  'J 

Problem  233.  Suppose  that  in  the  preceding  problem  the  boy 
weighs  65  Ib.  and  the  sled  10  Ib.,  and  that  the  boy  can  exert  a  force 

of  20  Ib.  horizontally  to  keep  him  on 
the  sled.  Will  the  boy  remain  on 
the  sled  when  the  latter  stops,  or 

will  he  be  thrown  forward  ?  f~. 
T 

Problem  234.       Two     weights, 

4j  G^lb.andG^lOlb.  (Fig.  157), 
attached  to  an  inextensible  cord 

#j  which  runs  over  a  pulley,  are  acted 
upon  by  gravity;  no  friction;  mo- 
tion takes  place.  Find  the  tension 
in  the  cord,  and  the  acceleration. 


Consider  G2  'and 


^  separately  with  the  forces  acting  upon  them,  and 
call   the  tension  in  the  cord  T. 
Then  apply  the  principle  "  acceler- 


citing    force 
acceleration.1 


equals 


mass     times 


FIG.  158 


Problem  235.  A  body  whose 
weight  G  =  5  Ib.  is  being  drawn 
up  an  inclined  plane  as  show  in 
Fig.  158  by  the  action  of  the 
weight  Gl  =  20  Ib.  Suppose  the 
resistance,  F,  offered  by  the  plane 
is  10  Ib.,  and  that  G  starts  from  rest.  How  far  up  the  plane  will  G 
go  in  6  sec.  ? 

Problem  236.  An  elevator  (Fig.  159),  whose  weight  is  2000  Ib.,  is 
descending  with  a  velocity,  at  one  instant,  of  2  ft.  per  second, 
and  one  second  later  it  has  a  velocity  of  18.1  ft.  per  second.  Find  the 


MOTION  IN  A    STRAIGHT  LINE 


191 


FIG.  159 


tension  T  in  the  cable  that  supports  the  elevator, 
assuming  uniform  retardation. 

Problem  237.  Suppose  the  elevator  in  preced- 
ing problem  going  up  with  the  same  acceleration. 
Find  the  tension  in  the  cable. 

Problem  238.  A  man  can  just  lift  200  Ib. 
when  standing  on  the  ground.  How  much  could 
he  lift  when  in*  the  moving  elevator  of  the  preced- 
ing problems,  (a)  when  the  elevator  was  ascend- 
ing? (b)  when  descending? 

Problem  239.      Two   weights,    G   and    G',   are 
connected  by  an  inextensible  flexible  cord  that  passes  over  a  friction- 
less  pulley,  as  shown  in  Fig.  160.     G  =  20  Ib.,  G'  =  100  Ib.,  and  there 

is  no  friction  on  the  plane. 

G  Find  the  tension  in  the  cord 

and  the  acceleration  of  the 
two  bodies. 

Problem  240.  A  30- 
ton  car  is  moving  with  a 
velocity  of  30  mi.  per  hour 
on  a  level  track.  The 
brakes  refused  to  work. 

How  far  will  the  car  go  after  the  power  is  turned  off  before  comin 
to  rest,  if  the  friction  is  .01  of  the  weight  of  the  car? 

109.  Variable  Acceleration.  —  It  has  already  been  shown 

dv 


FIG.  160 


,,  ds  dv      d2s 

that  v  =  — ,  «  =  —  =—, 

dt  dt      dt2 


and  a  =  v  —      These  relations 
ds 


hold  true  no  matter  whether  the  acceleration  is  constant 
or  variable.  If  the  acceleration  is  constant,  the  equations 
of  motion  are  those  that  have  already  been  worked  out 
(Art.  104),  and  by  simple  substitution  in  these  equations 
it  is  possible  to  find  the  velocity  in  terms  of  the  time,  the 


192  APPLIED  MECHANICS  FOR  ENGINEERS 

distance  in  terms  of  time,  and  the  distance  in  terms  of 
velocity. 

If  the  acceleration  is  variable,  it  is  necessary  to  work 
out  the  equations  of  motion  for  each  case.  This  may  be 
done,  when  it  is  known  how  a  varies,  by  means  of  either 
of  the  equations, 


or  vdv  =  ads. 

The  latter  equation  will  usually  give  the  beginner  less 
Difficulty. 

110.  Harmonic  Motion.  —  Let  it  be  supposed  that  a  body 
is  moved  by  an  attractive  force  which  varies  as  the  dis- 
tance. That  is,  the  attractive  force  is  proportional  to  the 
distance.  Then  the  acceleration  is  also  proportional  to 
the  distance. 

Let  the  acceleration        =  —  ks. 
Then  vdv  =  —  ksds, 

and  j    vdv  =  —  k\     sds  ; 

Jv0  J  o 

therefore  v2  —  v%  =  —  ks2, 

where  VQ  is  the  initial  velocity  when  8  equals  zero  and 
k  is  the  factor  of  proportionality,  deter  minable  in  any 
special  case.  This  equation  gives  the  relation  between 
the  velocity  and  distance.  Since  v  =  Vv§  —  &s2,  it  is 
evident  that  v  =  0  when  V&  •  s  =  VQ.  This  means  that  the 
body  comes  to  rest  when  s  has  reached  a  certain  value, 

viz.  ~~t'    From  the  original  assumption,  a  —  —  ks,  it  is 


MOTION  IN  A   STRAIGHT  LINE  193 

seen  that  the  acceleration  is  greatest,  when  s  is  greatest, 

# 

that  is,  when  s  =  — 7= ;  and  is  least  when  s  is  least,  that  is, 
vk 

when  s=0. 

To   get   the    relation  between  distance  and   time,   the 


equation  v  =  VtJJ  —  ks2  may  be  put  in  the  form 

ds 


from  which  —  —  sin"1  -  ~  =  t 


or 


sn 


This  relation  between  the  distance  and  time  shows  that 

i).  —  ^n 

as   t  increases  8  changes  in  value  from    —7=   to  —  ;=r,  as- 

V&          v& 

suming  all  values  between  these  limits,  but  never  exceed- 


ing them,  since  sin  V&£  can  never  be  greater  than  -f  1  or 
less  than  —1.  The  motion  is,  therefore,  vibratory  or 
periodic,  and  is  known  as  harmonic  motion.  The  complete 

•  A  •    *u-  -     27r 

period  in  this  case  is  —  z.* 

V* 

The  relation  between  velocity  and  time  may  be  found 
for  this  case  by  differentiating  the  last  equation  with 
respect  to  time.  Then, 

V    =    VQ    COS     VH. 

This  shows  that  VQ  is  the  greatest  value  of  v. 

This  motion  is  usually  illustrated  by  imagining  a  ball 
attached  to  two  pins  by  means  of  two  rubber  bands  or 


194  APPLIED  MECHANICS  FOR  ENGINEERS 

springs,  since  the  force  exerted  by  either  of  these  is  pro- 
portional to  the  elongation.  (See  Fig.  161.)  Assuming 
-  ^  that  there  is  no  friction  and 

[-. Q  [  that  the  ball  is  displaced  to 

£       o  a  position  B  by  stretching 

FIG>  161  one   of   the    rubber   bands, 

when  released  it  continues  to  move  backward  and  forward 
with  harmonic  motion. 

Problem  241.  Suppose  the  ball  in  Fig.  161,  held  by  two  helical 
springs,  to  have  a  weight  of  10  Ib.  and  that  it  is  displaced  1  in. 
from  0.  The  two  springs  are  free  from  load  when  the  body  is  at  O. 
The  springs  are  just  alike,  and  each  requires  a  force  of  10  Ib.  to 
compress  or  elongate  it  1  in.  Find  the  time  of  vibration  of  the 
body  and  its  velocity  and  position  after  |  sec.  from  the  time  when 
it  is  released. 

It  has  been  found  by  experiment  that  the  force  necessary  to  com- 
press or  elongate  a  helical  spring  is  proportional  to  the  compression 
or  elongation. 

111.  Motion  with  Repulsive  Force  Acting.  —  Suppose  the 
force  to  be  one  of  repulsion  and  to  vary  as  the  distance  ; 
then  a  =  Jcs,  and  vdv  =  ksds,  so  that 


v= 
s  = 


These  equations  show  that  as  t  increases  s  also  increases 
and  the  body  moves  farther  and  farther  away  from  the 
center  of  force.  The  motion  is  not  oscillatory. 

112.  Motion  where  Resistance  Varies  as  Distance.  —  If  a 
body  whose  weight  is  644  Ib.  falls  freely  from  rest  through 
60  ft.  and  strikes  a  resisting  medium  (a  shaft  where  fric- 


MOTION  IN  A    STRAIGHT  LINE 


195 


tion  on  the  sides  equals  2  F=  10  times  the 
distance  ;  see  Fig.  162),  since  accelerating 
force  equals  mass  times  acceleration, 


a  = 


a 

9 


=  9~ 


It  is  required  to  find  (a)  the  distance 
the  body  goes  down  the  shaft  before 
coming  to  rest ;  (£>)  the  distance  at  which 
the  velocity  is  a  maximum ;  (c)  the  total 
time  of  fall ;  (d)  the  velocity  at  a  distance 
of  10  ft.  down  the  shaft. 

After  striking  the  shaft  the  relation  be- 
tween velocity  and  distance  is  as  follows : 


FIG.  162 


The  remainder  of  the  problem  is  left  as  an  exercise  for  the 
student. 

Problem  242.  A  ball  whose  weight  is  32.2  Ib.  falls  freely  from 
rest  through  a  distance  of  10  ft.  and  strikes  a  400-lb.  spring  (Fig. 
163).  Find  the  compression  in  the  spring.  It  is  to  be  understood 
that  a  400-lb.  spring  is  such  a  spring  that  400  Ib.  resting  upon  it 
compresses  it  one  inch,  and  4800  Ib.  resting  on  it  compresses  it  one 
foot,  if  such  compression  is  possible.  After  the  ball  strikes  the  spring 
it  is  acted  upon  by  the  attraction  of  the  earth  and  the  resistance  of 

, ,        G  -  4800  5      , 

the  spring.  The  acceleration  a  is  then — ,  where  s  is  meas- 
ured in  feet.  The  relation  between  velocity  and  distance  is  then  ob- 
tained from  the  relation, 


196 


APPLIED  MECHANICS  FOE  ENGINEERS 


Q 


32.2 


Problem  243.  A  20-ton  freight 
car  (Fig.  164),  moving  with  a 
velocity  of  4  mi.  per  hour,  strikes 
a  bumping  post.  The  60,000-lb. 
spring  of  the  draft  rigging  of  the 
car  is  compressed.  Find  the  com- 
pression s.  Assume  that  the 
bumping  post  absorbs  none  of 
the  shock. 

Problem  244.  Suppose  the 
car  in  the  preceding  problem  to 
be  moving  with  a  velocity  of  4 
mi.  per  hour,  what  should  be 
the  strength  of  the  spring  in  the 
draft  rigging  so  that  the  com- 
pression cannot  exceed  2  in.  ? 

Problem    245.        After     the 
spring  in  Problem  242  has  been 
compressed  so  that  the  ball  comes 
to  rest,   it  begins  to  regain   its 
FIG.  163  original  form.      Find   the   time 

required  to  do  this  and  the  velocity  with  which  the  ball  is  thrown 
from  the  spring. 


4  Ml.  PER  HR. 


Q  = 


-1S\- 

WAMMAAD 


FIG.  164 


MOTION  IN  A   STRAIGHT  LINE 


197 


113.  Motion  when  Attractive  Force  Varies  Inversely  as  the 
Square  of  the  Distance.  —  This  is  the 
case  of  motion  (Fig.  165)  when  two 
bodies  in  space  are  considered,  since  in 
such  cases  the  attractive  force  varies 
directly  as  the  product  of  their  masses 
and  inversely  as  the  square  of  the  dis- 
tance between  their  centers  of  gravity. 
GThe  same  attraction  holds  between  two 
— opposite  poles  of  magnets  or  between 
two  bodies  charged  oppositely  with 
FIG.  165  electricity. 

—  k 
Suppose  the  acceleration  =  — —   and  that  the  velocity 

is  zero,  when  s  =  SQ. 


Then, 


so  that 


dt 


and     '     :          c_V£[VV-*-f^'*f  +  ?»} 

The  time  required  to  reach  the  center  of  attraction 
from  the  position  of  rest  is  obtained  by  putting  s  =  0. 

-   /,  \f 

frir  •  ,  7T     /  o,->  \ 

1  his  gives  t  =  — 

/T  V  9  / 

V&\^/ 

It  is  seen  that  when  s  =  0,  the  velocity  is  infinite,  and 
therefore  the  body  approaches  the  center  of  attraction 
with  increasing  velocity  and  passes  through  the  center,  to 
be  retarded  on  the  other  side  until  it  reaches  a  distance 
—  s0.  The  motion  will  be  oscillatory. 


198  APPLIED  MECHANICS  FOR  ENGINEERS 

If  one  of  the  bodies  is  the  earth,  of  radius  r,  and  the 
other  is  a  body  of  weight  Gr  falling  toward  it,  the  equa- 
tions just  derived  hold  true.  In  this  case  it  is  possible 
to  determine  k.  The  attraction  on  the  body  at  the  sur- 
face of  the  earth  is  6r,  and  at  a  distance  s  is  F,  so  that 

I*  =  Grf — \     The  acceleration  is  therefore  -^—  =  —  q(  —  \ 
\s2J  M  \s2J 

This  gives  &,  then,  equal  to  r^g. 

Substituting  these  values  in  the  above  equation,  we  find 


When  s  =  r  at  the  earth's  surface, 


If  «0  =  oo,  v  =  V2  gr. 

But  this  is  a  value  of  v  that  cannot  be  obtained,  since 
SQ  cannot  be  infinite.  So  that  the  velocity  is  always  less 
than  V2$r.  It  is  interesting  to  notice  here  that  if  a 
body  were  projected  from  the  earth  with  a  velocity  greater 
than  V2#r,  it  would  never  return,  provided  there  were 
no  atmospheric  resistance.  Substituting  (7  =  32.2  and 
r=3963  mi., 

v  =  6.95  mi.  per  sec. 

This  is  the  greatest  velocity  that  a  body  could  possibly 
acquire  in  falling  to  the  earth,  and  a  body  projected 
upward  with  a  greater  velocity  would  never  return 
(neglecting  resistance). 


MOTION  IN  A   STRAIGHT  LINE  199 

If  the  body  falls  to  the  earth  from  a  height  ^,  the  veloc- 
ity acquired  may  be  obtained  from  the  foregoing  by  put- 
ting s  =  r  and  «0  =  h  -f-  r  ;  then 


If  h  is  small  compared  to  r,  this  may  be  written,  without 
serious  error, 


which  is  the  formula  derived  for  a  freely  falling  body  in 
Art.  105. 

Problem  246.  A  body  of  weight  10  Ib.  has  an  initial  velocity  of  50 
f  /s  and  moves  in  a  medium  which  resists  with  a  force  proportional  to 
the  velocity  and  equal  to  1  Ib.  when  the  velocity  is  4  f  /s.  How  far 
will  the  body  move  before  the  velocity  is  reduced  to  10  f  /s?  Would 
the  velocity  ever  become  zero  with  that  law  of  resistance?  What 
would  be  the  limiting  value  of  the  space  passed  over  ? 

Problem  247.  A  body  of  weight  W  Ib.  is  projected  vertically 
upward  with  a  velocity  FQ.  If  the  resistance  of  the  air  is  equal  to  kv2, 
prove  that  the  height  to  which  the  body  will  rise  is 


Problem  248.     The  body  of  the  preceding  problem  falls  again  to 
the  earth.     Show  that  the  velocity  with  which  it  reaches  the  earth  is 


Problem  249.  A  man  jumps  from  a  balloon  and  acquires  a 
velocity  of  80  f /s  before  hi§  parachute  is  fully  opened.  The  man  and 
parachute  weigh  150  Ib.  The  resistance  of  the  air  varies  as  the 
square  of  the  velocity  (approximately)  and  is  1  Ib.  per  square  foot  of 
opposing  surface  when  the  velocity  is  20  f /s.  If  the  diameter  of  the 
parachute  is  12  ft.,  what  velocity  will  the  man  have  after  descending 
a  further  distance  of  200  ft.  ?  after  1000  ft.  ? 


200 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  250.  On  a  toboggan  slide  of  constant  slope  assume  the 
friction  to  be  constant  and  the  resistance  of  the  air  to  be  proportional 
to  the  square  of  the  velocity.  Derive  the  formula  for  the  velocity  in 
terms  of  the  distance,  starting  from  rest. 

114.  Relative  Velocity.  —  When  we  speak  of  the  velocity 
of  a  body,  it  is  understood  that  we  mean  the  velocity  of 
the  body  relative  to  the  earth,  more  particularly  the  point 
on  the  earth  from  which  the  motion  is  observed.  Since 
the  earth  is  in  motion,  it  is  evident  that  velocity  as  gen- 
erally spoken  of  is  not  absolute  velocity,  and  since  there 
is  nothing  in  the  universe  that  is  at  rest,  all  velocities 
must  be  relative.  In  everyday  life,  however,  we  think  of 
velocities  referred  to  any  point  on  the  surface  of  the  earth 
as  being  absolute. 

Suppose  two  particles,  A  and  B,  to  have  the  velocities 
Va  and  F"6,  as  illustrated  in  Fig.  166.  The  motion  of  B 
relative  to  A  is  obtained  by  regarding  A 
as  at  rest  and  B  as  having  a  velocity  com- 
posed of  the  actual  velocity  of  B  and  the 
reversed  velocity  of  A  at  each  instant. 
Thus,  the  vector  FJ  which  is  the  result- 
ant of  Vb  and  Va  reversed,  is  the  velocity 
of  B  relative  to  A  at  the  instant  at  which 
the  velocities  are  Va  and  Vb. 

This  is  made  more  evident  in  the  case 
of  constant  velocities  by  the  following 
consideration  :  if  Va  and  Vb  are  constant,  then  at  the  end 
of  a  unit  of  time  A  would  be  in  the  position  Av  and  B  in 
Bv  and  their  distance  apart  and  direction  from  each  other 
would  be  represented  by  the  line  A*BV  while  if  A  re- 


FIG.  166 


MOTION  IN  A   STRAIGHT  LINE 


201 


mained  at  rest,  and  B  moved  with  a  velocity  composed  of 
Vb  and  Va  reversed,  the  line  AB*  would  represent  the  dis- 
tance and  direction  of  B  from  A.  Since  AlBl  and  AB'  are 
evidently  equal  and  parallel,  the  relative  positions  of  the 
two  particles  would  be  the  same  in  one  case  as  in  the  other. 

As  an  illustration,  consider  the  velocity  of  the  wind 
relative  to  the  sail  of  an  ice  boat.     Let  Vb  and  Vw  be  the 
velocities     of    the 
boat  and  wind  re- 
spectively      (Fig. 
167)  and   SL  the 
direction     of     the 
sail,   which   is   as- 
sumed   to    be    a   - 
smooth  plane  sur- 
face. 

Combining  Vw 
with  Vb  reversed, 
the  velocity  V  is 
obtained,  which  is 
the  velocity  of  the 
wind  relative  to  the  boat.  This  velocity  can  be  resolved 
into  two  components,  OA  and  OE,  along  and  perpendicular 
to  the  sail  respectively.  The  component  OA  will  have  no 
effect  on  the  boat  on  the  assumption  that  the  sail  is  a 
smooth  surface.  The  component  OE  may  be  resolved  into 
two  components,  OD  and  00,  respectively,  along  and  per- 
pendicular to  the  path  of  the  boat.  The  vector  OD  then 
represents  the  component  of  the  velocity  of  the  wind  which 
urges  the  boat  forward. 


FIG.  167 


202  APPLIED  MECHANICS  FOB   ENGINEERS 

From  the  figure  it  is  clear  that  there  may  be  a  forward 
component  of  the  wind's  velocity  on  the  boat  even  when 
the  velocity  of  the  boat  is  greater  than  the  velocity  of  the 
wind.  It  is  only  necessary  for  the  velocity  of  the  wind 
relative  to  the  boat  to  fall  in  front  of  the  sail.  Then  as 
long  as  the  forward  pressure  of  the  wind  is  greater  than 
the  resistance  of  the  ice  the  velocity  of  the  boat  will 
increase.  In  the  case  of  ice  boats,  where  the  resistance 
is  small,  the  velocity  of  the  boat  may  greatly  exceed  the 
velocity  of  the  wind  for  high  wind  velocities. 

Problem  251.  A  train  is  moving  with  a  speed  of  60  mi.  per 
hour,  and  another  train  on  a  parallel  track  is  going  in  the  opposite 
direction  with  a  speed  of  40  mi.  per  hour.  What  is  the  velocity 
of  the  second  train  as  observed  by  a  passenger  on  the  first  ? 

Problem  252.  A  man  in  an  automobile  going  at  a  speed  of  40 
mi.  per  hour  is  struck  by  a  stone  thrown  by  a  boy.  The  stone  has 
a  velocity  of  30  ft.  per  second  and  moves  in  a  direction  perpendicular 
to  the  direction  of  motion  of  the  automobile.  With  what  velocity 
does  the  stone  strike  the  man? 

Problem  253.  A  man  attempts  to  row  across  a  river,  \  mi.  wide, 
which  is  flowing  at  the  rate  of  3  mi.  per  hour.  If  he  can  row  at  the 
rate  of  4  mi.  per  hour,  what  direction  must  he  take  in  rowing  in  order 
to  reach  a  point  directly  across  on  the  opposite  shore.  What  distance 
will  he  row  relative  to  the  water?  How  long  will  it  take  for  him  to 
cross? 

Problem  254.  An  ice  boat  is  moving  due  north  at  a  speed  of  60 
mi.  per  hour,  and  the  wind  blows  from  the  southwest  with  a  velocity 
of  40  mi.  per  hour.  What  is  the  apparent  velocity  and  direction  of  the 
wind  as  observed  by  a  man  on  the  boat  ? 

Problem  255.  A  man  walks  in  the  rain  with  a  velocity  of  4 
mi.  per  hour.  The  raindrops  have  a  velocity  of  20  ft.  per  second  in  a 
direction  making  60°  with  the  horizontal.  How  much  must  the  man 
incline  his  umbrella  from  the  vertical  in  order  to  keep  off  the  rain  :  (a) 


MOTION  IN  A   STRAIGHT  LINE  203 

when  going  against  the  rain,  (&)  when  going  away  from  the  rain  ?  If 
he  doubles  his  speed,  what  change  is  necessary  in  the  inclination  of  his 
umbrella  in  (a)  and  (6)  V 

Problem  256.  The  light  from  a  star  enters  a  telescope  inclined 
at  an  angle  of  45°  with  the  surface  of  the  earth.  The  velocity  of 
light  is  186,000  mi.  per  second  and  the  earth  (radius  4000  mi.)  makes 
one  revolution  in  24  hr.  What  is  the  actual  direction  of  the  star  with 
respect  to  the  earth  ?  This  displacement  of  light  due  to  the  velocity 
of  the  earth  and  the  velocity  of  light  is  known  as  aberration  of  light. 

Problem  257.  A  locomotive  is  moving  with  a  velocity  of  40  mi.  per 
hour.  Its  drive  wheels  are  80-in.  in  diameter.  What  is  the  tangential 
velocity  of  the  upper  point  of  the  wheels  with  respect  to  the  frame  of 
the  locomotive?  What  is  the  tangential  velocity  of  the  lowest  point? 

Problem  258.  Show  that  it  an  ice  boat  were  moving  north  and 
the  wind  had  a  velocity  of  20  mi.  per  hour  from  the  southwest,  the 
maximum  velocity  that  the  boat  could  attain  if  there  were  no  friction 
would  be  38.6  mi.  per  hour  if  the  sail  were  set  at  30°  with  the  direction 
of  the  boat's  motion.  What  effect  would  increasing  the  angle  between 
the  sail  and  the  boat's  direction  have  in  this  case  ? 

Problem  259.  With  the  sail  set  at  30°  with  the  direction  of  the 
boat's  motion,  in  what  direction  could  the  boat  go  fastest  if  there 
were  no  friction?  Ans.  When  the  direction  of  the  boat  makes  an 
angle  of  60°  with  the  direction  of  the  wind. 

Problem  260.  Given  that  the  angle  which  the  sail  makes  with 
the  direction  of  motion  of  the  ice  boat  is  30°,  the  angle  which  the 
direction  of  the  wind  makes  with  the  direction  of  the  boat's  motion 
is  45°,  the  area  of  the  sail  is  50  sq.  ft.,  the  resistance  of  the  ice 
to  the  boat's  motion  is  16  lb.,  and  given  that  the  pressure  of  the 
wind  normal  to  the  surface  of  the  sail  varies  as  the  square  of  the 
velocity  (component  of  relative  velocity  normal  to  sail),  and  is  1  lb. 
per  square  foot  of  opposing  surface  when  the  velocity  is  15  mi.  per 
hour,  find  the  maximum  velocity  that  the  boat  can  attain  when  the 
velocity  of  the  wind  is  (a)  20  rni.  per  hour,  (5)  30  mi.  per  hour. 

Ans.    (a)  14.6  mi.  per  hour,     (b)  34.0  mi.  per  hour. 


CHAPTER   X 


dt 
FIG.  168 


CURVILINEAR  MOTION 

115.  Velocity.  —  Suppose  a 
particle  to  be  moving  along 
the  curve  of  Fig.  168,  from 
A  toward  B.  The  average 
velocity  of  the  particle  be- 
tween A  and  B  is  defined  as 
the  velocity  the  particle 
would  have  if  it  moved  uni- 
formly from  A  to  B,  i.e. 
along  the  chord  A  B  with  con- 
stant speed.  The  average 
velocity  of  the  particle  between  A  and  B  is  therefore, 

average  velocity  = ^-^ 

time  spent  between  A  and  B 

If  the  chord  is  represented  by  Ac  and  the  time  by  A£, 
then 

average  velocity  =  — , 

and  is  represented  by  a  vector  A  C,  of  length  — ,  laid  off 
from  A  on  AB.  When  the  time  A£  is  made  to  approach 
the  limit  zero,  the  limiting  value  of  the  ratio  —  is  defined 
to  be  the  velocity  of  the  particle  at  the  point  A. 

204 


CURVILINEAR  MOTION  205 

The  limiting  direction  of  the  chord  is  the  tangent  at  A. 
Hence,  calling  v  the  velocity  of  the  particle  at  the  point  A, 

do 

v  =  — 
dt 

and  is  represented  by  a  vector  of  length  —  laid  off  from 

dt 

A  along  the  tangent  in  the  direction  of  motion. 

It  is  shown  in  calculus  that  if  a  single  tangent  to  the 
curve  at  A  exists,  then  the  limit  of  the  ratio  of  the  arc  to 
its  chord,  as  the  arc  approaches  the  limit  zero,  is  unity. 
Hence,  if  the  arc  AB  is  As,  then 

ds  _  -, 
dc~ 

and  v  =  ^  =  ^.. 

dt  dc      dt 

The  velocity  of  a  particle  moving  along  a  curved  path  is 
therefore  represented  at  any  point  of  the  path  by  a  vector 

equal  to  the  value  of  —  laid  off  on  the  tangent  to  the  path 
dt 

at  the  given  point,  and  in  the  direction  of  the  motion. 

116.  Acceleration  in  a  Curved  Path. — Let  the  velocities 
at  A  and  B  (Fig.  169  (a))  be  v  and  v  +  Av  respectively. 
Lay  off  the  vectors  representing  these  velocities  from  the 
same  point  A  (Fig.  169  (5)). 

The  vector  (7Z>,  from  the  end  of  v  to  the  end  of  v  H-  At;, 
represents  the  change  in  velocity  in  the  time  A£.  The 
ratio  of  this  vector  to  the  time  A£  is  the  average  accelera- 
tion for  the  time  A*.  Thus  the  vector  CE  represents  the 
average  acceleration  of  the  particle  between  A  and  B. 


206  APPLIED  MECHANICS  FOR  ENGINEERS 

The  limiting  value  of  this  average  acceleration  as  A£ 
approaches  zero  as  a  limit  is  defined  to  be  the  acceleration 
of  the  particle  at  A.  Letting  vx  and  vy  be  the  projections 


(b) 


FIG.  169 


of  v  upon  rectangular  axes,  then  the  projections  of  CD 
on  these  axes  are  Avx  and  Avtf  and  the  projections  of  CE 
are  and  ^  Hence 


and  calling  a  the  acceleration  of  the  particle  at  A, 


If  ax  and  ay  are  the  components  of  a  along  the  axes,  then 


d,Vr 
a«=^f'         «> 


dt 


Now 


vx  =  v  cos  6  and  vy  =  v  sin  6 


CURVILINEAR  MOTION 


207 


where  0  is  the  angle  which  the  tangent  to  the  curve  at  A 
makes  with  the  #-axis.     Therefore 

/i  dv          .     *  dO  .    fidv  ,  ndO 

ax  =  cos  0 v  sin  0  — ,  ay  =  sm  6  —  +  v  cos  6  —  • 

dt  dt       y  dt  dt 


Since 


_<fo     dv  _  d?s 
~~~       ~~ 


dO 


A  value  of  —  may  be  found  as  follows  :  Draw  a  normal 
dt 

to  the  curve  at  A  (Fig.  170).     On  this  normal  there  is  a 
point  (9',  the  center  of  a  circle  tangent  to  the  curve  at  A 
and  passing  through  B.     Letr'  be  the  radius  of  this  circle 
and   A0'    the   angle 
subtended  at   0'  by  o' 

the  arc  AB.  As  B 
is  made  to  approach 
A  as  a  limit  the  tan- 
gents to  the  curve 
and  the  circle  at  B 
take  the  same  limit- 
ing position  and 

hence  the  ratio  — — 
A0 

approaches  1  as  a  limit.  The  ratio  of  the  arcs,  As'  of  the 
circle,  and  As  of  the  curve,  has  the  limit  1,  since  they  have 
the  same  chord  and  the  limit  of  the  ratio  of  each  arc  to  its 
chord  is  1. 

As  the  point  B  is  made  to  coincide  with  A  the  point  Of 
takes  some  limiting  position,  0,  which  is  called  the  center 
of  curvature  of  the  curve  at  A.  If  r  is  the  value  of  OA, 
then  limit  rf  =  r. 


FIG.  170 


208  APPLIED  MECHANICS  FOR  ENGINEERS 


Now  r'A0'=A«',   or         =     . 

Ar      H 

rp,  ,  d#'          1  ^0  1  ,      df« 

Therefore  —  -  =  -,  or  since  —  —  =  1  and  —  -  =  1, 

ds'      r  dd'  ds' 

—  =  -    and—  =  —  —  =-  —  =  -. 
ds      r  '  dt      ds  dt     r  dt      r 

The  quantity  r  is  the  radius  of  curvature  of  the  curve  at  A. 

Substituting  the  values  found  for  —  and  —  ,  the  values 
of  ax  and  av  become 


117.  Tangential  and  Normal  Components  of  the  Accelera- 
tion. —  The  above  values  of  ax  and  ay  hold  for  any  rectangu- 
lar axes.  Let  the  #-axis  be  taken  along  the  tangent  at  A 
and  the  y-axis  along  the  normal.  Then  cos  0=1,  and 
sin  6  =  0  at  the  given  point,  and  the  values  of  ax  and  ay  be- 
come respectively 

d*S  V* 

at  =  dT*>    an=r' 

The  normal  force  and  tangential  force  may  now  be 
written, 

Normal  force  =  ^?J 
r 

Tangential  force  =M^g* 
civ 

For  all  curves  except  the  circle  r,  the  radius  of  curvature 
varies  from  point  to  point.  In  the  circle,  however,  it  is  the 
radius,  and  is  therefore  constant.  In  this  case  the  normal 
force  is  usually  called  the  centripetal  force, 


i*tt 


CURVILINEAR  MOTION 


a-v, 


OAQ 

A\JU 


Problem  261.  A  ball  weighing  2  Ib.  is  attached  to  one  end  of  a 
string  5  ft.  long,  the  other  end  of  which  is  attached  to  a  fixed  point. 
The  ball  is  pulled  aside  and  released.  When  the  string  makes  an  angle 
of  30°  with  the  vertical,  the  velocity  of  the  ball  is  10  f/s.  Find  the 
normal  and  tangential  components  of  the  acceleration  and  the  accel- 
eration in  magnitude  and  direction  at  that  instant.  Find  also  the  ten- 
sion in  the  string.  (Notice  that  the  force  acting  toward  the  center  on 
the  body  is  the  difference  between  the  tension  in  the  string  and  the 
component  of  the  weight  along  the  direction  of  the  string.) 

118.     Uniform  Motion  in  a  Circle. — A  body  of  mass  M 

moving  with  constant  speed,  v,  in  the  circumference  of  a 

V2 

circle  of  radius  r  has  an  acceleration  —  directed  toward  the 

•  r 

center  of  the  circle.     The  force  necessary  to  impart  this 
acceleration  to  the  body  is  — — ,  also  directed  toward  the 

center.  Hence  this  force  must 
be  the  resultant  of  all  the  forces 
acting  on  the  body.  An  illustra- 
tion of  uniform  motion  in  a  circle 
is  the  simple  governor  shown  in 
Fig.  171.  When  the  speed  is 
constant,  then  a,  A,  and  r  are  con- 
stant. Let  T  be  the  tension  in 
the  rod  supporting  the  ball,  then, 
since  there  is  no  vertical  motion 
=  0,  so  that  T  cos  «=  G. 


FIG.  171 


Considering  the  normal  force,  we  have  T  sin  a 


Mv* 


so 


that  tan  a  =  —     From  these  equations   T  may  be  found 
gr 

for  any  values  of  a  and  r. 


210 


APPLIED  MECHANICS  FOE  ENGINEERS 


G   =  10  LB8. 


Problem     262.        The 

weighted  governor  shown 
in  Fig.  172  is  rotated  at 
such  a  speed  that  a  =  30°. 
Find  the  forces  acting  on 
the  longer  rods  and  the 
stress  in  the  shorter  rods. 
The  connections  are  all 
pin  connections. 

Problem  263.  A  type 
of  swing  is  shown  in  Fig. 
173.  A  revolving  central 
post  supported  by  wires  A 
and  B  carries  six  cars  G, 
each  suspended  from  cross 
arms  D  by  means  of  cables 
50  ft.  long.  When  the 
swing  is  at  rest,  the  cars 
hang  vertically  and  a  —  0 ; 
as  the  speed  of  rotation  increases,  ct,  becomes  larger.  Suppose  the  car 
and  its  load  of  four  passengers  to  weigh  1000  lb.,  and  the  speed  to  be 
such  that  a]=  30° ;  find  the  tension  in  the  cables  supporting  the  cars. 
Assume  that  a  single  car  is  carried  by  one  cable. 

Problem  264.  The 

same  principle  that 
has  been  seen  to  hold 
for  motion  in  a  circle 
enables  us  to  solve  a 
problem  that  comes 
up  in  railroad  work. 
When  a  train 


FIG.  172 


around  a  curve,  it  is 
desirable  to  have  the 
outer  rail  raised  suffi- 
ciently so  that  the 


CURVILINEAR  MOTION 


211 


wheel  pressure  will  be  normal  to  the  rails.  It  is  really  the  same 
problem  as  Problem  263,  where  the  sustaining  cable  is  replaced  by 
a  track.  (See  Fig.  1740  Let  r  be  the  radius  of  curvature,  v  the  veloc- 
ity of  the  car,  of  weight  G.  Show  that  the  superelevation  of  the 

outer  rail  is  given  by  tan  a  =  — ,  and 

gr 

so,  approximately, 

7.      ^2 
h  =  — , 

9r 

where  d  is  the  distance  between  the 
rails  in  feet,  v  the  velocity  in  feet  per 
second,  g  is  32.2,  r  is  the  radius  of 
curvature  in  feet,  and  h  is  the  superelevation  of  the  outer  rail  in  feet. 

Problem  265.     Show  that,  using  d  =  4.9  ft.,  this  height  may  be 
expressed,  approximately,  as  follows  : 


where  h  and  r  are  in  feet  and  i\  is  the  velocity  in  miles  per  hour. 

Problem  266.  Find  the  superelevation  of  the  outer  rail  on  a 
curve  of  radius  2000  ft.  for  speeds  of  20  mi.  per  hour,  40  mi.  per  hour, 
60  mi.  per  hour. 

Problem  267.  What  would  the  superelevation  need  to  be  for  the 
above  speeds  on  a  curve  of  radius  500  ft.  ? 

119.  Motion  without  Friction  along  Any  Curve  in  a  Vertical 

Plane.  —  Let  a 
particle  slide 
down  the  smooth 
curved  track 
(Fig.  175),  hav- 
ing a  velocity  VQ 
at  the  point  A 
FlG.  175  distant  yQ  above  a 


212  APPLIED  MECHANICS  FOR  ENGINEERS 

horizontal  line  of  reference.  When  in  the  position  P,  the 
forces  acting  on  the  particle  are  the  weight  G,  acting  ver- 
tically downward,  and  the  force  exerted  by  the  track, 
which  is  normal  to  the  track  since  there  is  no  friction. 
The  tangential  force  is  then  Cr  sin  6,  and  hence  for  de- 
scending motion, 

G-sm0=—  at. 
9 

But  at  =  0—  -,  and  ds  •  sin  0  =  —  du 
ds 

where  s  is  the  length  of  the  curve  AP  measured  from  A 
toward  P. 


Therefore  I    vdv  =  —  g  I    dy, 

«/V0  •JVO 


or 


Hence,  the  change  in  velocity  acquired  by  a  particle  in 
sliding  down  a  smooth  track  is  the  same  at  any  point  as  if  the 
particle  had  fallen  freely  through  the  same  vertical  distance. 
Compare  Art.  105. 

The  time  of  descent  depends  upon  the  form  of  the 
curve. 

Problem  268.     Show  that  the  equation 

vdv  =  —  gdy 

holds  when  the  body  is  ascending  as  well  as  when  descending,  and 
then  prove  that  the  body  would  rise  on  the  track  to  the  level  from 
which  it  started  from  rest. 


CURVILINEAR  MOTION 


213 


Problem  269.  A  body  of  weight  G  starts  from  rest  at  the  top  of 
a  smooth  circular  track  of  radius  R  in  a  vertical  plane.  Find  the 
velocity,  acceleration,  and  force  exerted  by  the  track  on  the  body 
when  it  has  traveled  through  60°,  90°,  150°,  180°,  210°  of  arc. 

Problem  270.  With  what  velocity  would  a  body  have  to  be  pro- 
jected from  the  lowest  point  of  a  smooth  circular  vertical  track  in 
order  that  its  velocity  would  just  keep  it  from  falling  from  the  track 
at  the  top? 

Problem  271.  If  the  body  of  the  preceding  problem  were  pre- 
vented from  leaving  the  track,  what  velocity  at  the  lowest  point 
would  just  carry  it  around  the  track? 

Problem  272.  In  the  centrifugal  railway  (Fig.  176),  friction 
neglected,  what  would  have  to  be  the  ratio  of  h  to  h'  so  that  the  car 
would  not  leave  the  track  at  A,  starting  from  rest  at  the  height  h  ? 


FIG.  176 

Problem  273.  If  the  car  in  the  preceding  problem  were  prevented 
from  leaving  the  track,  what  would  be  the  ratio  of  h  to  h'  to  just 
carry  the  car  past  A  ? 

Problem  274.  A  particle  slides  from  rest  from  any  point  on  a 
smooth  sphere.  Show  that  it  will  leave  the  sphere  when  it  has  de- 
scended vertically  through  one  third  of  its  original  vertical  distance 
above  the  center. 


214 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  275.  A  locomotive  weighing  175  tons  moves  in  an  800- 
ft.  curve  with  a  velocity  of  40  mi.  per  hour.  Find  the  horizontal  pres- 
sure on  the  rails,  if  they  are  on  the  same  horizontal. 

Problem  276.  If  the  velocity  of  the  earth  was  18  times  what  it 
actually  is,  show  that  the  force  of  gravity  would  not  be  sufficient  to 
keep  bodies  on  the  earth  near  the  equator.  Take  the  radius  of  the 
earth  as  4000  mi.,  and  assuming  the  above  conditions,  find  at  what 
latitude  the  body  would  just  remain  on  the  earth. 

Problem  277.  A  pail  containing  5  Ib.  of  water  is  caused  to  swing 
in  a  vertical  circle  at  the  end  of  a  string  3  ft.  long.  Find  the  velocity 
of  the  pail  at  the  highest  point  so  that  the  water  will  remain  in  the 
pail.  Find  also  the  velocity  of  the  pail  at  the  lowest  point. 

120.  Simple  Circular  Pendulum.  —  The  simple  circular 
pendulum  consists  of  a  weight  Or  suspended  by  a  string 

without  weight,  of 
length  Z,  in  such  a  way 
that  it  is  free  to  move 
in  a  circle  in  a  vertical 
plane  due  to  the  action 
of  gravity.  (See  Fig. 
177.)  Let  E  be  such  a 
position  of  the  weight 
that  its  height  above 
the  lowest  position  0  is 
—  A,  and  Q  any  other  po- 
sition designated  by  the 
coordinates  x  and  y. 
Let  the  weight  be  6r 

and  the  tension  in  the  string  T.  These  are  the  only  forces 
acting  on  the  body.  The  only  forces  that  can  produce 
motion  in  the  circle  are  those  that  are  tangent  to  the  circle 


FIG.  177 


CURVILINEAR  MOTION  215 

OB.  The  force  T  is  normal  to  the  circle  and  so  has  no 
tangential  component.  The  force  6r  has  a  tangential  com- 
ponent —  6r  sin  6.  The  equation  of  motion  is,  therefore, 

vdv  =  ads  =  —  g  sin 


where  0  =  —  -  =  -,    where  s   denotes    distance   along 
i  i 

the  curve  measured  from  0. 

Since  s  =  W,  ds  =  Id6, 
and  hence         vdv  —  —  lg  sin  BdO.  (1) 

Approximate  Solution.  The  integration  of  this  equa- 
tion leads  to  a  complicated  relation  between  the  angle 
and  the  time.  An  approximate  solution  can  be  obtained 
for  small  angles  of  oscillation  by  replacing  sin  6  by  6. 
Equation  (1)  then  becomes 

vdv  =  -lg9dO. 

Integrating,  v2  =  —  lg(fi  +  0. 

Let     v=   Owhen   0=^;  then  C=lgO\. 


For  the  body  descending  8  decreases  as  t  increases  and 

therefore  —  -  is  negative,  and 
at 


S^'-Jjta 

-02  *J 


Integrating,    sin-1—  =  —  \f *+  Gv 
r,  I 


216  APPLIED  MECHANICS  FOR  ENGINEERS 

Let  t  =  0  when  Q—Q^  then  sin"1!  =  (7r 
From  the  angles  whose  sines  are  1  choose  ^=  sin~1l=  Or 


Then  V~  •  <  =  -  -  sin-1^-.  (2) 

»   7  9  /5 

1 

As  the  pendulum  swings  from  the  highest  point  on  the 
right  to  the  point  on  the  left  on  the  same  level  (Art.  119), 
i.e.  to  where  0  =  —O^  the  value  of  t  continually  increases. 

Equation  (2)  shows  that  sin"1—  must  continually  decrease 

fi    e_ 

during  this  time.  Therefore  sin  1^  must  decrease  in  this 
interval  from  —  to  sin"1  (  —  1),  which  therefore  can  only 

be  —  — .     Hence  if  t  is  the  time  of  a  single  oscillation, 


or 


This  value  of  t  is  called  the  period  of  vibration. 
Q-eneral  Solution.     For  large  vibrations  the  above  ap 
proximate  solution  is  not  sufficiently  accurate. 
Integrating  equation  (1), 

vdv  =  —  gl  sin  6d6, 

we  obtain  ^-  =  gl  cos  0  +  0. 

2 

When  v  =  0,  0  =  0V 

.-.  C=  —  gl  cos  01 
and  v*=2l   cos  0  —  cos  0 . 


CURVILINEAR  MOTION 


217 


=  -  V20Z  (cos  0- cos  0 


Vcos  0  —  cos  Bl 
Since  cos  0  =  1-2  sin2  ^,  this  equation  may  be  written 


If  ^  is  the  time  that  it  takes  for  the  pendulum  to  descend 
from  0  =  6l  to  0  =  0,  then 


or 


The  time  of  vibration,  «,  is  double  this,  or 


Change  the  variable  by  the  substitution 


Then 
and 


.0       .    6,    .  •  . 
sm-  =  sm-^sm</ 

"1/5  /3 

-  cos  -  dO  =  sin  — 1  cos 

22  2 


*  = 


d<f> 


218  APPLIED  MECHANICS  FOR  ENGINEERS 

This  integral  is  an  elliptic  integral.  The  expression  can  be 
integrated  only  by  expanding  into  a  series  and  integrating 
term  by  term. 

t\ 
Thus,  letting  Jc  =  sin  -1, 

2 


From  the  reduction  formula 

f  sin-  xdx  =  -  cosa;sin"'la;  +  2^1  fsin-»  xdx 
J  n  m    J 

there  follows 

/*2.  w-lp.  ,   _(w 

I  smm  ^a;  =  --  I  sinw~2  xdx  - 
»/Q  ^  «/o 

and  hence 


where  &  =  sin  -! 


Problem  278.  Using  the  series,  compute  correct  to  three  figures 
the  time  of  vibration  of  a  simple  pendulum  of  length  3  ft.  when  the 
pendulum  swings  through  20°.  Compare  with  the  time  obtained 

from  t  =  ir\— 
* 


CUEVILINEAE  MOTION  219 

Problem  279.  Compute  the  time  of  vibration  of  a  simple  pendu- 
lum of  length  4  ft.  when  swinging  through  an  angle  of  80°. 

Problem  280.  A  pendulum  vibrates  seconds  at  a  certain  place  and 
at  another  place  it  makes  60  more  vibrations  in  12  hours.  Compare 
the  values  of  g  at  the  two  places. 

Problem  281.  Show  that  in  the  simple  pendulum  for  any  angle 
of  swing  the  value  of  the  tension  in  the  string  is 


where  h  and  y  have  the  meaning  given  them  in  Art.  120. 

Hence  show  that  the  tension  is  equal  to  the  weight  when  the 
weight  has  descended  through  one  third  the  original  vertical  distance. 

Problem  282.  The  weight  of  a  chandelier  is  300  lb.,  and  the  dis- 
tance of  its  center  of  gravity  from  the  ceiling  is  16  ft.  Neglecting 
the  weight  of  the  supporting  chain,  find  how  much  the  tension  in  the 
chain  will  be  increased  if  the  chandelier  is  set  swinging  through  an 
angle  of  8°,  measured  at  the  ceiling. 

121.  Cycloidal  Pendulum.  —  It  has  been  found  that  a 
pendulum  may  be  obtained  whose  period  of  vibration  is 
constant  by  allowing  the  string  to  wrap  itself  around  a 
cycloid  as  shown  in  Fig.  178.  The  pendulum  hangs  from 
the  point  A.  AB  and  A  O  are  cycloidal  guides  around 
which  the  string  wraps  as  the  pendulum  swings.  This 
causes  the  length  of  the  pendulum  to  continually  change 
and  the  pendulum  "bob  "  to  move  in  another  cycloidal 
curve  COB.  The  equation  of  this  curve  referred  to  the 
axes  x  and  y  is 

x  =  -  vers     f -*j  -f  \  -f  -  #2. 

In  Art.  119  it  was  seen  that  v2  =  %  g  (A—  y)  represented 
the  velocity  of  a  body  moving  in  a  vertical  curve  when 


220  APPLIED  MECHANICS  FOR  ENGINEERS 

only  the  force  of  gravity  and  a  force  normal  to  the  path 
of  the  curve  acted.     We  may  make  use  of  the  equation  in 
r 


FIG.  178 

this  case,  since  the  same  conditions  exist.     We  may  write 


ds 

,    since  v  =  — - 
at 


From  the  equation  of  the  curve,  we  find 


so  that 


taking  the  negative  sign,  since  t  is  a  decreasing  function 
of    . 


CURVILINEAR  MOTION  221 


Therefore        t  =  \/—  Tvers"1 
M#[_ 


J0 

The  whole  time  of  vibration  is  twice  this  value,  so  that 
the  time  of  vibration 


This  expression  is  independent  of  A,  so  that  all  vibrations 
are  made  in  the  same  time.  The  motion  is  therefore 
isochronal. 

Problem  283.     Using  the  calculus  formula  for  radius  of  curva- 
ture, 


dx* 

show  that  the  radius  of  curvature  of  the  cycloid  of  the  above  article 
at  any  point  is 


and  hence  show  that  the  tension  in  the  string  for  any  position  is 


where  G  is  the  weight  of  the  pendulum  bob. 

Problem  284.     Assuming  a  value  of  h  in  terms  of  /,  say  h  =  - , 

plot  the  curve  representing  the  tension  in  terms  of  y.      For  what 
position  of  the  pendulum  is  the  tension  a  maximum  ? 
Show  that  for  any  swing  the  maximum  tension  is 

T  = 


Problem  285.     A    particle  slides    from   rest  down  an  inverted 
cycloid.     Prove  that  its  vertical  velocity  is  greatest  when  it  has  de- 


222  APPLIED  MECHANICS  FOR  ENGINEERS 

scended  through  one   half  the  original  vertical  distance  above  the 
lowest  point  of  the  curve.     What  is  the  vertical  velocity  at  that  point? 

122.  Motion  of  Projectile  in  Vacuo.  —  A  method,  slightly 
different  from  the  preceding,  of  dealing  with  a  problem 

of  curvilinear 
motion,  is  il- 
lustrated in 
the  present  ar- 
ticle. It  is  de- 
X  sired  to  find 
the  path  taken 


Fia-  179  by  a  body  pro- 

jected with  a  velocity  VQ  at  an  angle  of  elevation  a,  when 
the  resistance  of  the  air  is  neglected.  (See  Fig.  179.) 
Let  the  point  of  projection  be  taken  as  origin  and  the  z-axis 
horizontal.  Then,  since  there  is  no  horizontal  force  act- 
ing on  the  body,  a.  =  0,  so  that 

^  =  0 
& 

dx 

and  —  =  constant  =  v0  cos  a,  ^^ 

therefore  x  =  v0cos  a(t). 

In  a  similar  way  we  know  that  the  vertical  acceleration 
ay  =  —  #,  since  the  only  force  acting  is  G. 

Then, 


and  — ^  =  —  gt  +  constant. 

at 


CURVILINEAR  MOTION  223 

This  equation  may  be  rewritten 

vy  =  —  gt  +  constant. 

To  determine  this  constant  of  integration,  we  put  t  =  0, 
and  vv  =  VQ  sin  a  ; 

therefore  -^  =  -  gt  +  v0  sin  a 

at 


and  y  =  -  J  #*2  +  v0  sin  «(0  • 

Eliminating  £  between  the  equations  in  x  and  ^,  we  get 


2  Vl  COS2  a 

as  the  equation  of  the  path  of  the  projectile.     This  is  evi- 
dently a  parabola,  with  its  axis  vertical. 

Range.     To  find  the  range  or  horizontal  distance  d  we 
put  y  =  0  ;  then  x  =  0  and  x  —  d, 

so  that  d  = 


From  this  it  is  clear  that  the  greatest  range  is  given  when 

^^v  2 

a  =  45°,  since  then  d  =  — . 

The  Greatest  Height.     The  greatest  height  to  which  the 
projectile  will  rise  is  found  by  putting  x  =   °  -  in 

t/ 

the  equation  of  the  curve  and  solving  for  y.     This  gives 

v2Q  sin2  a 

h  ==  — ~ } 

%g 

and  the  angle  that  gives  the  greatest  height  is  a  =  90°. 

2 

For  this  case  h  =  — -.     This  is  the  case  that  has  already 


224 


APPLIED  MECHANICS  FOE  ENGINEERS 


been  considered  under  the  head  of  a  body  projected  verti- 
cally upward. 

Problem  286.  A  fire  hose  delivers  water  with  a  nozzle  velocity  ^o? 
at  an  angle  of  elevation  a.  How  high  up  on  a  vertical  wall,  situated 
at  a  distance  d'  from  the  nozzle,  will  the  water  be  thrown  ?  It  should 
be  said  that  water  thrown  from  a  nozzle  in  a  non-resisting  medium 
takes  a  parabolic  path  and  follows  the  same  laws  as  projectiles. 

Problem  287.  What  at  least  must  be  the  nozzle  velocity  of  water 
thrown  upon  a  burning  building,  200  ft.  high,  the  angle  of  elevation 
of  the  nozzle  being  60°? 

Problem  288.  The  muzzle  velocity  of  a  gun  is  500  ft.  per  second. 
Find  its  greatest  range  when  stationed  on  the  side  of  a  hill  which 
makes  an  angle  of  10°  with  the  horizontal :  (a)  up  the  hill,  (6)  down 
the  hill.  If  the  hillside  is  a  plane,  prove  that  the  area  commanded 
by  the  gun  is  an  ellipse,  of  which  the  gun  is  a  focus. 


FIG.  180 


CURVILINEAR  MOTION 


225 


Problem  289.     From  the  foot  of  a  plane  inclined  ft  to  the  hori- 
zontal a  projectile  is  fired  at  an  angle  «  to  the  horizontal  up  the  plane. 
Prove  that  the  range  on  the  inclined  plane  is 
2  vl  cos  a.  sin  (a  —  /?) 


Show  that  the  range  is  a  maximum  when  a  =  45°  +  —  • 

Problem  290.  Prove  that  the  initial  velocity  v0  is  the  same  as 
that  of  a  body  falling  freely  from  the  directrix  of  the  parabolic  path 
to  the  point  0  on  the  curve.  Show  that  the  velocity  of  the  body  at 
any  point  on  the  curve  is  the  same  as  would  be  acquired  in  falling 
freely  from  the  directrix  to  that  point.  (See  Fig.  179.) 

Problem  291.  A  ball  whose  weight  is  64.4  lb.,  shown  in  Fig.  180, 
starts  from  rest  at  A  and  slides  without  friction  in  a  circular  path  to 
the  point  B,  where  it  is 
projected  from  the  cir- 
cular path  horizontally. 
Find  (a)  the  velocity 
at  5,  (6)  the  equation 
of  its  path  after  leaving 

.B,  and  (c)  the  distance  / 

d  from  a  vertical 
through  B,  where  it 
strikes  a  horizontal  10 
ft.  below  B. 


0 


Problem  292.        If 

the  body  in  Problem 
291  had  moved  along 
a  straight  line  from  A 
to  -B  and  was  then  pro- 
jected, find,  as  in  the 
preceding  problem,  (a), 
(6),  and  (c). 

Problem  293.     A  body  whose  weight  is  12  lb.  swings  as  a  circular 
pendulum,  as  shown  in  Fig.  181,  from  A  to  B,  when  the  string  breaks. 
Q 


FIG.  181 


226  APPLIED  MECHANICS  FOR  ENGINEERS 

Find  (a)  the  velocity  at  B,  (b)  the  equation  of  its  path  after  leaving 
B,  and  (c)  the  distance  d  where  it  strikes  a  horizontal  5  ft.  below  B. 

Problem  294.  The  muzzle  velocity  of  a  gun  situated  at  a  height  of 
300  ft.  above  a  horizontal  plane  is  2000  ft.  per  second.  Find  the  area 
of  plane  covered  by  the  gun. 

Problem  295.  A  projectile  is  fired  from  a  mortar  gun  with  an 
initial  velocity  of  25  mi.  a  minute.  What  is  the  maximum  range 
on  the  horizontal,  neglecting  air  resistance  ? 

123.  Motion  of  Projectile  in  Resisting  Medium.  —  It  was 
found  by  Rollins  and  others  (see  Encyclopaedia  Britannica 
—  "Gunnery  ")  that  the  formula  for  projectiles  in  vacuo  did 
nothold  when  the  projectile  moved  in  the  atmosphere.  That 
is,  that  the  path  followed  by  the  projectile  was  not  para- 
bolic, but  on  account  of  the  resistance  of  the  atmosphere 
the  range  was  much  less  than  that  given  by  the  parabola. 

A  formula  constructed  by  Helie,  empirically  modifying 
the  parabolic  formula,  is 


2 


where  &=  0.0000000458  —  ,  d  being  the  diameter  of  the 

w 

projectile  in  inches,  and  w  its  weight  in  pounds.  This 
gives  the  simplest  formula  for  roughly  constructing  a 
range  table. 

Professor  Bashforth  of  Woolrich  found,  from  a  series 
of  experiments  made  by  him,  that  for  velocities  between 
900  and  1100  ft.  per  second  the  resistance  varied  as  v6,  for 
velocities  between  1100  and  1350  ft.  per  second  the  re- 
sistance varied  as  v3,  and  for  velocities  above  1350  ft.  per 
second  the  resistance  varied  as  v2. 


CURVILINEAR  MOTION 


227 


In  addition  to  the  resistance  of  the  air  other  factors  tend 
to  change  the  path  of  the  projectile  from  the  parabolic 
form,  viz.  the  velocity  of  the  wind  and  the  rotation  of  the 
projectile  itself.  Most  projectiles  are  given  a  right-handed 
rotation,  and  this  causes  them  to  bear  away  to  the  right 
upon  leaving  the  gun.  This  is  called  drift.  Correction 
is  made  for  drift  and  wind  velocity  upon  firing. 

Problem  296.  Taking  VQ  =  1000  f /s,  a  =  45°,  d  =  6  in.,  w  =  150  lb., 
find  the  range  from  Helie's  formula.  Plot  to  the  same  scale  the  curve 
followed  by  the  projectile  and  the  parabola  it  would  follow  if  there 
were  no  resistance. 

Problem  297.  With  the  projectile  of  the  preceding  problem  and 
v0  =  1200  f/s,  find  the  angle  of  elevation  to  strike  a  point  200ft.  high, 
distant  1000  ft.  horizontally,  (a)  using  HeUie's  formula,  (6)  using  the 
parabola. 

124.  Motion  in  a  Twisted  Curve.  —  If  a  particle  moves 
along  a  curve  in  space,  it  follows,  as  in  the  case  of  a  two- 
dimensional  curve,  that  the  ve- 
locity of  the  particle  at  any  point 
of  the  curve  is  in  the  direction  of 
the  tangent  to  the  curve  at  that 
point,  and  its  value  is 


<     .        .          ,  =  .      (Fig.  182.) 

If   the   tangent  makes   angles   «, 

$,   7,    with   the  coordinate   axes, 
then 


= 


v  cos 


vg 


FIG.  182 


dz 

—  =  v  cos  7. 

dt 


If  a  is  the  acceleration  of  a  particle  when  at  the  point  P 


228  APPLIED  MECHANICS  FOR  ENGINEERS 

and  ax,  ay,  az,  the  components  of  a  parallel  to  the  axes,  then 


d(y  cos  /3) 


_  d2^  _  c?(t;  cos  7) 

^~ 


dv    •   da 
or  ax  =  cos  a  —  —  v  sin  a  —  , 


dt  dt' 

a._co87|-«in7|. 

The  sum  of  the  projections  of  a#  ay,  ag  on  the  tangent 
is  the  component  of  the  acceleration  along  the  tangent;  i.e. 

at  =  ax  cos  a  +  av  cos  (3  +  a,  cos  7 
=  (cos2  a  +  cos2  /3  +  cos2  7)  — 

—  v[  sin  a  cos  a  —  +  sin  /3  cos  ft  —  +  sin  7  cos  7  -j-  \ 

\  a£  dt  dt/ 

But  cos2  a  +  cos2  ft  -f  cos2  7  =  1,  and  therefore 

—  2[  cos  a  sin  a  ~  +  cos  /3siup-@-  +  cos  7  sin  7  -^  ]  =  0. 

\  dt  dt  dt/ 

Therefore  at~~dt 

If  a  particle  slides  without  friction  down  any  space 
curve,  it  can  be  shown  as  in  the  corresponding  case  of  the 


CURVILINEAR  MOTION 


229 


plane    curve    (Art.    109)    that   the   velocity   attained   in 
descending  a  vertical  distance  h  from  rest  is  given  by 


In  the  actual  case  of  a  body  sliding  along  a  curved 
chute,  friction  on  the  side  and  bottom  of  the  chute  and 
air  resistance  reduce  the  velocity  below  that  given  by  the 
above  formula. 

As  an  illustration  consider  the  motion  of  a  body  along  a 
helical  chute  with  vertical  axis  (Fig.  183). 


If  the  z-axis  is  the  axis  of  the  helix  and  the  curve  passes 
through  a  point  on  the  rs-axis,  then 

x  =  r  cos  #, 
y  =  r  sin  6, 

*  =  A0, 

27T 

where  d  is  the  pitch  of  the  helix,  or  vertical  distance 
traveled  by  the  generating  point  per  turn.  Developing 
one  thread  of  the  helix,  it  is  seen  that  the  tangent  makes  a 
constant  angle  7  with  the  z-axis  such  that 


230  APPLIED  MECHANICS  FOR  ENGINEERS 

During  the  motion  of  the  body  the  chute  exerts  a  force 

<\ 

towards  the  axis  of  the  curve  of  value  —-  •     This  force 

r 

causes  a  retarding  friction  force  along  the  tangent  of  value 


c      -,  where  c  is  a  constant  depending  upon  the  roughness 

of  surface  of  chute  and  body.  In  addition  there  is  a  tan- 
gential friction  force  due  to  the  force  exerted  by  the  bot- 
tom of  the  chute  on  the  body.  The  component  of  the 
weight  perpendicular  to  the  bottom  of  the  chute  is  IF  sin  7 
and  the  friction  force  caused  by  it  is  c  W  sin  7.  Therefore 
the  total  retarding  force  is 


gr  J 

The  component  of  the  weight  along  the  tangent  is  IF  cos  7. 
Hence  the  net  accelerating  force  along  the  tangent  is 

/  vv 

Ft=W[  cos  7  —  c  sin  7  —  c  — 

\  ; 

=  — -(b2  —  v2),   where  b2  =  ^-cos  7  —  gr  sin  7. 
gr  c 

Hence  the  acceleration  along  the  tangent  is 


or 


Therefore 


Integrating,    J-  log    ±^  =  c-t  +  C\ 


—  v     r 


CURVILINEAR  MOTION  231 

If  the  body  starts  from  rest,  v  =  0  when  t  =  0. 


Therefore  -  =  e~r~  , 

b  —  v 

Solving  for  v, 


2bct 

^~  +  1 


This  formula  shows  that  v  remains  always  less  than  the 
constant  quantity  b. 

On  account  of  the  velocity  not  exceeding  a  fixed  value 
this  form  of  chute  is  useful  in  sending  packages  from  one 
floor  of  a  building  to  another. 

Problem  298.  Given  a  helical  chute  of  radius  3  ft.  and  pitch  2  ft., 
find  the  velocity  of  a  small  body  descending  from  rest  after  5  sec., 
after  10  sec.,  if  the  value  of  c  is  .05. 


FIG.  184 


CHAPTER   XT 
ROTARY  MOTION 

125.  Angular  Velocity  and  Angular  Acceleration  of  a  Par- 
ticle. —  If  a  particle  moves  in  any  plane  curve,  the  rate  at 
which  a  line  joining  the  particle  to 
a  fixed  point  in  the  plane  is  turning 
is  called  the  angular  velocity  of  the 
particle  with  respect  to  that  point. 

Thus,  if  6  is  the  angle  which 
the  line  joining  the  particle  makes 
with  a  fixed  line  through  0  (Fig. 
184),  the  angular  velocity,  o>,  of  the  particle  with  respect 
to  Ois 

<*  =  di' 

The  rate  of  change  of  the  angular  velocity  is  called 
the  angular  acceleration  of  the  particle  with  respect  to  the 
point. 

Thus,  if  a  is  the  angular  acceleration  of  the  particle 
with  respect  to  0, 

The  angular  acceleration  may  also  be  written  in  the  form 
da  dO  d<* 


«   vt 


ROTARY  MOTION  233 

The  angular  velocity  and  angular  acceleration  of  a 
particle  moving  in  a  given  path  with  a  given  speed  de- 
pend upon  the  point  to  which  they  are  referred. 

Problem  299.  Prove  that  if  a  particle  moves  around  a  circle,  the 
angular  velocity  of  the  particle  with  respect  to  a  point  on  the  circum- 
ference is  equal  at  any  instant  to  one  half  the  angular  velocity  of  the 
particle  with  respect  to  the  center  of  the  circle,  and  hence  that  the 
angular  velocity  of  the  particle  at  any  instant  is  the  same  with  re- 
spect to  all  points  on  the  circumference. 

Show  also  that  a  like  statement  holds  for  angular  acceleration. 

Problem  300.  Show  that  if  a  particle  have  constant  angular 
acceleration  with  respect  to  any  point,  the  following  formulae  hold : 


2. 


•*-«5  =  2a0, 

where  6  is  the  angle  turned  through  in  the  time  t,  o)0  the  initial  an- 
gular velocity,  and  <o  the  angular  velocity  at  the  instant  t.  Compare 
these  formulae  with  those  of  Art.  104. 

Problem  301.  A  flywheel  making  100  revolutions  per  minute  is 
brought  to  rest  in  2  min.  Find  the  angular  acceleration  a  and  the 
angle  0  turned  through  before  coming  to  rest. 

Problem  302.  A  flywheel  is  at  rest,  and  it  is  desired  to  bring  it 
to  a  velocity  of  300  radians  per  minute  in  \  min.  Find  the  angular 
acceleration  ct  necessary  and  the  number  of  revolutions  required. 
What  is  the  angular  velocity  o>  at  the  end  of  10  sec.  ? 

126.  Motion  in  a  Circle.  —  If  a  particle  is  moving  in  a 
circle,  the  linear  velocity  at  any  instant  is 

«  =     .  (Art.  115.) 


234 


APPLIED  MECHANICS  FOR   ENGINEERS 


But  here  *  =  rd, 

where  s  is  the  arc  passed  over  and  6  the  angle  subtended 

by  this  arc  at  the  center. 

ds  =  rd0 
'  dt     T  dt" 

Or  V  :~~  f*tA 

Taking  the  second  derivative, 


dt* 


From  Art.  117  —  is  the  tangen- 
d/v 

tial  component,  at,  of  the  linear 
acceleration  of  the  point,  and  hence 

at  =  r<x. 

Problem  303.     Prove  that  for  a  particle  moving  in  a  circle  with 
angular  velocity  <o  about  the  center, 


FIG.  185 


dt 


=  Xto), 


where  x  and  y  are  the  coordinates  of  the  position  of  the  particle  re- 
ferred to  rectangular  axes  through  the  center. 

Problem  304.  The  balance  wheel  of  a  watch  goes  backward  and 
forward  in  f  sec.  The  angle  through  which  it  turns  is  180°.  Find 
the  greatest  angular  acceleration  and  the  greatest  angular  velocity, 
given  that  the  angular  acceleration  varies  directly  as  the  angular 
displacement  and  is  toward  the  neutral  position  of  the  spring. 

Problem  305.  Suppose  the  flywheel  in  Problem  302  to  be  6  ft. 
in  diameter.  After  arriving  at  the  desired  angular  velocity,  what  is 
the  tangential  velocity  of  a  point  on  the  rim?  What  has  been  the 
tangential  acceleration  of  this  point,  if  constant? 


7 


ROTARY  MOTION  235 

Problem  306.  Assume  that  the  angular  acceleration  of  a  particle 
varies  inversely  as  the  square  of  the  angle  turned  through;  find 
the  relation  between  CD  and  0,  and  t  and  6. 

127.  Angular  Velocity  with  Respect  to  an  Axis.  —  If  a  par- 
ticle moves  along   any  space  curve,  its  angular  velocity 
with  respect  to  any  fixed  axis  is 

the  rate  at  which  a  line  passing 
through  the  particle  and  the  axis, 
perpendicular  to  the  axis,  is  turn- 
ing about  the  axis  ;  or  it  is  the 
angular  velocity  of  the  projection 
of  the  moving  particle  on  a  plane 
perpendicular  to  the  axis  with 
respect  to  the  foot  of  the  axis  on  that  plane  (Fig.  186). 

Problem  307.     The  coordinates  of  a  moving  particle  are 
x  =  r  cos  ct,   y  =  r  sin  ct,   z  =  kt, 

where  r,  c,  and  k  are  constants  and  t  is  the  time.     Prove  that  the 
angular  velocity  about  the  z-axis  is  c.    What  is  the  curve  traced  out  ? 

128.  Plane  Motion  of  a  Body.  —  A  body  is  said  to  have 
plane  motion  when  each  point  of  the  body  moves  in  a  fixed 
plane.     The  plane  in  which  the  center  of  gravity  of  the 
body  moves  will  be  called  the  plane  of  motion. 

The  rate  of  turning  of  any  fixed  line  of  the  body  in  the 
plane  of  motion  is  called  the  angular  velocity  of  the  body. 
It  should  be  noted  that  the  idea  of  rotation  about  an  axis 
does  not  enter  into  this  definition  of  angular  velocity  of  a 
body. 

129.  Relation  between  the  Velocities  of  Points  of  a  Body 
having  Plane  Motion.  —  Let  A  and  B  (Fig.  187)  be  two 


236  APPLIED  MECHANICS  FOR  ENGINEERS 

T          ^~       ~^>v    ..  points  of  the  body  in  the 

plane  of  motion.  Choose 
axes  OX  and  OY  in  the 
plane  of  motion. 

If  (x,  y)  and  (d ',  ?/')  are 
the  coordinates  of  A  and  B 
respectively  and  6  is  the 
angle  which  AB  makes 
with  the  #-axis,  then 
rcostf,  (1) 

(2) 


FIG.  187 


x'  = 


where  r  is  the  constant  distance  AB. 


Therefore 


dxf     dx          •    adO 

——•     (    __     A»     QT  Y\      M     __^ 

dt       dt  dt ' 


dt 


(3) 

(4) 


dx 


If  A  were  a  point  fixed  in  space,  —  and  -^  would  be 

dt  dt 

zero   and   the    components   of   the  velocity  of   B  would 
become 


dt 


, 

dt 


dt 


dt 


If  A  is  in  motion,  ~  and  -Jf  are  the  components  of  the 
dt  dt 

velocity  of  A.  It  therefore  follows  from  equations  (3)  and 
(4)  that  the  velocity  of  B  is  compounded  of  the  velocity  of 
A  and  the  velocity  of  B  relative  to  A  regarded  as  at  rest. 


EOTARY  MOTION 


237 


Hence  if  va  is  the  velocity  of  A  at  any  instant  and  a> 
the  angular  velocity  of  the  body  at  that  instant,  the  veloc- 
ity of  B  is  the  resultant  of  va  and 
ra),  the  latter  being  at  right  angles 
to  AB  (Fig.  187). 

Problem  308.  A  wheel  of  radius  3  ft.  rolls 
along  a  straight  line,  the  velocity  of  the  center 
being  10  f/s  (Fig.  188).  Find  the  velocities 
of  the  points  A,  B,  D,  and  E  shown  in  the 
figure.  Find  the  velocity  of  A  relative  to 
B,  of  E  relative  to  A,  of  D  relative  to  E. 

Problem  309.     Show  that  the  velocity  of  B  in  Fig.  187  is 


FIG.  188 


vb  =  Vt£  +  rV2  +  2  ra>va  sin  0.         f\  - 

130.  Relation  between  the  Accelerations  of  Points  of  a  Body 
having  Plane  Motion.  —  Taking  the  derivatives  of  the 
equations  (3)  and  (4)  of  Art.  129,  it  follows  that  the  ac- 
celeration of  a  point  in  the  plane 
of  motion  of  a  body  having  plane 
motion  is  composed  of  the  accel- 
eration of  any  other  point  in  the 
plane  of  motion  and  the  accelera- 
tion of  the  first  point  relative  to 
the  second  regarded  as  at  rest. 

Thus  if  a  body  in  plane  motion 
has  angular  velocity  co  and  angular  acceleration  «,  the  accel- 
eration ab  of  a  point  B  is  made  up  of  #,  the  acceleration  of 
A,  ro)2  along  BA,  and  ret,  perpendicular  to  AB  (Fig.  189). 

Problem  310.  Find  the  acceleration  of  each  of  the  points  A,  B, 
D,  E,  in  Problem  308,  given  that  the  velocity  of  the  center  is  con- 
stant. Represent  these  accelerations  by  vectors. 


Fia.  189 


238 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  311.  If  the  point  C  in  Problem  308  has  an  acceleration 
of  2  f/s2  in  the  direction  of  motion  at  the  given  instant,  find  the 
acceleration  of  each  of  the  points  A,  B,  D,  E. 

Problem  312.  Show  that  for  a  wheel  rolling  along  a  straight 
line  the  tangential  velocity  and  acceleration  of  any  point  of  the  cir- 
cumference relative  to  the  center  regarded  as  at  rest  are  the  same  in 
magnitude  as  the  velocity  and  acceleration  of  the  center  of  the  wheel. 

Problem  313.  A  locomotive  drive  wheel  6  ft.  in  diameter  rolls 
along  a  level  track.  Find  the  greatest  tangential  acceleration  and 
the  greatest  normal  acceleration  of  any  point  on  the  tread,  (a)  when 
the  constant  velocity  v  with  which  the  wheel  moves  along  the  track 
is  60  mi.  per  hour,  (6)  when  the  engine  is  slowing  down  uniformly 
and  has  a  velocity  of  30  mi.  per  hour  at  the  end  of  3  min.,  (c)  when 
the  engine  is  starting  up  uniformly  and  has  a  velocity  of  30  mi.  per 
hour  at  the  end  of  5  min. 

131.  Instantaneous  Axis  of  Rotation  for  Plane  Motion.  — 
Let  A  and  B  be  two  points  of  a  body  having  plane  mo- 
tion, lying  in  the  plane  of  motion. 
If  the  body  does  not  have  a  motion 
of  translation  only,  A  and  B  can 
always  be  chosen  so  that  their 
directions  of  motion  are  not  par- 
allel. Through  A  draw  in  the 
plane  of  motion  a  line,  AO,  per- 
pendicular to  the  direction  of 
motions  of  A.  Any  point  of  the 
body  on  AO,  if  it  is  in  motion  at 

all,  must  be  moving  perpendicular  to  AO,  since  any  two 
points  of  the  body  remain  always  the  same  distance  apart. 
Likewise,  a  point  on  OB,  which  is  perpendicular  to  the 
direction  of  motion  of  B,  must,  if  moving  at  all,  be  moving 


FIG.  190 


ROTARY  MOTION  239 

perpendicular  to  OB.  The  point  of  intersection,  0,  of 
OA  and  OB,  must  therefore  be  at  rest,  since  it  cannot  at 
the  same  time  be  moving  perpendicular  to  OA  and  OB. 

The  line  through  0,  perpendicular  to  the  plane  of  mo- 
tion, is  called  the  instantaneous  axis  of  rotation.  Every 
point  on  this  axis  has  velocity  zero  at  the  given  instant, 
and  the  motion  of  the  body  is  at  the  instant  one  of  rota- 
tion about  this  line. 

If  the  figure  in  motion  is  a  plane  figure  moving  in  the 
plane  in  which  it  lies,  the  point  0  is  called  the  instanta- 
neous center  of  rotation. 

The  path  traced  by  the  instantaneous  center  is  called 
the  centrode. 

If  OA  =  rv  OB  =  r2,  and  the  velocities  of  A  and  B  are 
respectively  vl  and  v2,  then  the  angular  velocity  of  the 

body  is  v       v 

<»  =  -i  =  -2. 

rl        r2 

It  follows  that  v1:vz  =  r1:  rz. 

Problem  314.  A  straight  line  of  length  a  moves  with  its  ends  on  two 
axes  at  right  angles  to  each  other.  Prove  that  the  centrode  is  a  quadrant 
of  a  circle  of  radius  a  and  center  at  the  intersection  of  the  two  axes. 

Problem  315.     A  connecting  rod  is  4  ft.  long,  and  the  crank-pin 
circle   is   1  ft.  in  radius.      Sketch  the 
centrode  of  the  connecting  rod  for  one 
complete  stroke. 

Problem  316.  The  diameter  of  the 
wheel  outlined  in  Fig.  191  is  6  ft.  The 
velocities  of  the  points  B  and  D  are  as 
indicated  by  the  vectors.  Find  (a)  the 
instantaneous  center,  (6)  the  magnitude  FIG.  191 


240 


APPLIED  MECHANICS  FOE  ENGINEERS 


of  the  velocity  of  the  point  D,  (c)  the  angular  velocity  of  the  wheel, 
(d)  the  velocity  of  the  point  E,  (e)  the  velocity  of  slipping  of  the 
point  A  of  the  wheel. 

132.  Simultaneous  Rotation  of  a  Body  about  Two  Intersect- 
ing Axes.  —  Suppose  a  body  rotating  about  an  axis  OA 
(Fig.  192)  with  an  angular  velocity  co^  relative  to  a  frame- 

y 

I>/*<*2 

JE 


£<&r 


work  attached  to 
OA,  and  let  OA 
and  the  attached 
framework  be  at  the 
same  instant  rotat- 
ing about  an  axis 
OB  with  angular 
velocity  ®2. 

If  P  be  a  point  of 
the  body,  or  a  point  of  space  thought  of  as  moving  with 
the  body,  in  the  plane  containing  OA  and  OB,  between 
OA  and  OB,  and  distant  r^  and  r2,  respectively,  from  these 
lines,  the  velocity  of  P  due  to  the  two  rotations  is 


v  = 


FIG.  192 


If  P  is  so  chosen  that  r^  and  r2  satisfy  the  relation  r± :  r2 
=  e»2  :  cov  the  velocity  of  P  is  zero.  The  locus  of  all  such 
points  is  clearly  a  straight  line  through  0.  Hence  there 
is  a  straight  line,  OP,  in  the  plane  of  the  two  intersecting 
axes  of -rotation  which  is  at  rest  at  the  given  instant,  and 
about  which  the  body  is  rotating  at  that  instant. 

To  find  the  angular  velocity,  o>,  of  the  body  about  this 
instantaneous  axis  OP,  consider  the  motion  of  the  point 


EOTAEY  MOTION 


241 


0  on  OA.     Draw  OE  perpendicular  to   OB  and  OF  per- 
pendicular to  OP. 

The  point  C  has  angular  velocity  o>2  about  OB,  and 
hence  its  velocity  is  ECa^.  Its  angular  velocity  about 
OP  is  ft)  and  hence  its  velocity  is  FQ(o. 


Now 


=  0(7  sin  a  and  ^(7=  0(7  sin  0. 


•.  a)  sin  a  =  ft>2  sin  0. 


Again,  FC=r1cosa^  and  EO=r^-\- 
Therefore  ^a)  cos  a  =  r2co2  +  r-^co^  cos  0. 

But  r2co2  =  rjttfj, 

and  hence  co  cos  a  =  co1  +  o)2  cos  0. 

Squaring  and  adding  equations  (1)  and  (2), 

«2  =  (Of  +  w|  -f   2  «ift)2  COS  0. 

Dividing  (1)  by  (2), 

tan  a  =  — ^2  sin . 

Wj  +  ft)2  COS  0 


cos 


0. 


(2) 


(3) 


(4) 


Equations   (3)   and   (4)   are  exactly  the  equations  that 
would  be  obtained  by  re- 
garding coj  and  ft)2  as  vec- 
tors and  finding  their  re- 
sultant as  in  Fig.  193. 

Hence  we  may  represent 
angular  velocity  of  a  body 
about  a  line  by  a  vector 
laid  off  on  that  line,  equal 
in  length  to  the  numerical  ° 
value  of  the  angular  ve- 


242  APPLIED  MECHANICS  FOR  ENGINEERS 

locity,  and  may  combine  simultaneous  angular  velocities 
about  two  intersecting  lines  by  combining  their  vectors  in 
the  usual  way.  The  resultant  vector  represents  the  value 
of  the  resultant  angular  velocity  and  coincides  with  the 
instantaneous  axis  of  rotation  of  the  body. 

The  vector  representing  an  angular  velocity  is  made 
to  point  opposite  to  the  direction  from  which  the  rota- 
tion appears  counter-clockwise  to  an  observer  looking 
along  the  axis  of  rotation. 

It  is  evident  at  once,  then,  that  the  angular  velocity 
of  a  body  about  a  line  may  be  regarded  as  made  up  of 
component  angular  velocities  about 
two  or  more  axes  intersecting  that 
line,  obtained  by  using  the  parallelo- 
gram law  of  vectors.     In  particular 
a  rotation  about  an  axis  may  be  re- 
garded as  made  up  of  the    simulta- 
^-  r  I         neous  rotations  about  three  axes   at 

/  "v _/          right  angles  to  each  other  and  inter- 

FIG.  194  secting  the  given  line  (Fig.  194). 

If  the  angular  velocity  of  the  body 

about  the  line  is  ft),  the  angles  which  this  line  makes  with 
the  three  rectangular  axes,  a,  fi,  7,  and  the  angular  velocities 
of  the  body  about  these  axes  ft)x,  coy,  coz  respectively,  then 

ft)x  =  ft)  cos  a,         coy  =  co  cos  /3,         ft)z  =  ft)  cos  7, 
and  co2  =  ft)J  -|-  o)2  +  &)2. 

Problem  317.  Prove  that  the  vector  representation  holds  for 
simultaneous  rotation  about  two  parallel  axes,  i.e.  that  the  resultant 
angular  velocity  of  Wj  and  tu2  is 


ROTARY  MOTION 


243 


and  that  the  instantaneous  axis  of  rotation  divides  the  line  joining 
the  axes  of  Wj  and  w2  in  the  inverse  ratio  of  o^  and  o>2.     (See  Fig.  195.) 

Problem  318.  A  wheel  is  making  120  r.  p.  m.  about  a  shaft  and 
the  shaft  at  the  same  time  is  making  90  r.  p.  m.  about  a  line  perpen- 
dicular to  the  axis  of  the  shaft.  Find  the  instantaneous  axis  of  the 
shaft  at  any  instant. 


-e- 


A 

0 

B 

\ 

^  fcj. 

8" 

C 

1 

fi 

D 

FIG.  195 


FIG.  196 


Problem  319.  A  body  has  simultaneous  rotations  o^  =  4  TT  rad./sec., 
o>2  =  3  TT  rad./sec.,  about  the  parallel  lines  of  Fig.  196  in  the  direc- 
tions indicated.  Locate  the  instantaneous  axis  of  rotation  and  find 
the  angular  velocity  of  the  body  about  that  axis. 

133.  Rotation  of  the  Earth.  Foucault's  Pendulum.  —  Let 
a)  be  the  angular  velocity  of  the 
earth  about  its  axis  of  rotation 
OP  (Fig.  197).  This  angular 
velocity  can  be  resolved  into  two 
components  about  rectangular 
axes  CA  and  OB.  If  X  is  the 
latitude  of  the  position  of  A^  the 
component  of  G>  about  CA  is  o> 
sin  X.  If  a  plane  containing  CA 
could  be  fixed  so  as  not  to  turn 
about  CA,  the  rotation  of  the  earth  about  CA  would  be 
indicated  by  the  apparent  rotation  of  the ,  plane  in  the 


FIG.  197 


244  APPLIED  MECHANICS  FOR  ENGINEERS 

opposite  direction  about  0!A,  since  our  directions  are 
measured  with  reference  to  the  earth's  surface. 

Such  a  plane  is  the  plane  of  vibration  of  a  heavy  weight 
suspended  by  a  long  wire. 

Foucault  was  the  first  to  demonstrate  the  rotation  of 
the  earth  by  this  method. 

Problem  320.  Show  that  the  rotation  of  the  earth  about  a  di- 
ameter through  a  point  on  the  equator  is  zero. 

Problem  321.     Show  that  the  time  for  the  plane  of  the  Foucault's 

pendulum  to  turn  through  360°  is hours,  where  X  is  the  latitude 

sin  A 
of  the  place. 

Problem  322.  The  plane  of  a  Foucault's  pendulum  is  observed 
to  turn  through  26°  in  2  hr.  20  min.  Find  the  latitude  of  the  place. 


CHAPTER   XII 
WORK  AND  ENERGY 

134.  Definitions.  —  If  a  force,  constant  in  magnitude 
and  direction,  acts  at  a  definite  point  of  a  body,  the  force  is 
said  to  do  work  on  the  body  when  the  point  of  application 
of  the  force  has  a  displacement  with  a  component  in  the 
direction  in  which  the  force  acts. 

The  product  of  the  force  and  the  component  of  the  dis- 
placement in  the  direction  of  the  force  is  defined  to  be 
the  work  done  on  the  body  by  the  force  in  that  displace- 
ment. If  there  is  a  displacement  opposite  to  the  direction 
of  the  force,  work  is  said  to  be  done  against  the  force. 

For  example,  in  Fig.  198,  let  the  forces  P,  R,  6r,  JVact 
upon  the  block  as  the  block  moves  from  the  position  A  to 


FIG.  198 

the  position  B.  If  BC  is  perpendicular  to  the  line  of  P, 
the  displacement  in  the  direction  of  P  is  AO  and  the 
work  done  by  P  on  the  block  is  P  -  A  O.  If  AB  =  s,  then 
AC—  s  cos  0,  and  the  work  done  by  P  is  P«  cos  0.  This 

245 


246  APPLIED  MECHANICS  FOR  ENGINEERS 

may  be  written  P  cos  6  -  s.  But  P  cos  0  is  the  component 
of  P  in  the  direction  of  the  displacement  of  the  point  of 
application,  and  therefore  the  work  done  by  P  is  equal 
to  the  product  of  the  displacement  of  the  point  of  appli- 
cation and  the  component  of  the  force  in  the  direction  of 
the  displacement. 

Since  N  and  G-  are  at  right  angles  to  the  displacement, 
no  work  is  done  by  -ZVor  (7. 

Since  the  displacement  is  opposite  to  the  direction  of  R, 
work  is  done  against  jR,  the  amount  being  Rs. 

If  the  body  had  moved  from  B  to  A  during  the  action 
of  the  given  forces,  the  work  done  by  R  would  be  jRs,  and 
the  work  done  against  P  would  be  P  cos  0  -s . 

If  the  point  of  application  changes  in  the  body,  a  dif- 
ferent definition  of  work,  of  which  the  above  definition  is 
a  special  case,  may  be  given. 

Definition :  The  rate  at  any  instant  at  which  work  is 
being  done  by  a  force  acting  on  a  body  is  the  .product  of 
the  velocity  of  the  particle  of  the  body  on  which  the  force 
is  acting  at  the  given  instant  and  the  component  of  the 
force  in  the  direction  of  the  velocity  of  that  particle. 

For  example,  in  Fig.  199  suppose  a  force  P,  applied  to 

the  block  B,  drags  B 

o  ^  I       I  ^  g   '         ^] 

s  1  [^          '\B        p     along  A  and  at  the 

—* *  '  '    ff*.     same  time  moves  A. 

Let  F  be  the  for- 
ward force  exerted 
by  B  on  A.  If  when  A  moves  forward  a  distance  *, 
B  moves  relative  to  A  a  distance  «',  then  the  rate  at 
which  work  is  being  done  by  F  acting  on  block  A  is 


WORK  AND  ENERGY  247 

F  •  —  ,  and  the  rate  at  which  work  is  being  done  against 
dt 

.F  acting  on  block  B  is  F  .d(s  +  *').     Hence  the  rate  at 

ctt 

tf 

which  work  is  being  done  against  the  forces  of  resistance 
between  the  bodies  is 


dt  dt         '   dt 

and  the  work  thus  done  in  the  time  t  in  which  B  slips  on 

Cl    d&r  Cs* 

A  the  distance  s'  is  I  F—  -  dt  or  I    FdJ. 
Jo      dt  Jo 

If  F  is  constant  this  reduces  to  Fs'. 

135.  Friction  Forces.  —  Wherever  the  surfaces  of  two 
bodies  come  in  contact  and  there  is  motion  of  one  body 
along  the  other,  or  any  force  tending  to  produce  such 
motion,  each  body  exerts  upon  the  other  a  force  along 
their  common  surface,  a  force  tending  to  prevent  the  rela- 
tive motion  of  one  surface  along  the  other.  This  resist- 
ing force  is  known  as  friction.  The  laws  of  friction  are 
discussed  in  a  later  chapter.  In  general  the  law  of  fric- 
tion of  unlubricated  surfaces  may  be  expressed  by  saying 
that  the  limiting  force  of  friction  for  two  given  surfaces 
when  slipping,  or  just  ready  to  slip,  on  each  other  is  pro- 
portional to  the  normal  pressure  between  the  surfaces. 

The  ratio  of  this  limiting  force  of  friction  to  the  normal 
force  is  called  the  coefficient  of  friction. 

From  the  preceding  article  it  is  seen  that  the  work 
done  against  the  friction  F  between  two  surfaces  is  equal 
to  \  Fds,  or  in  case  F  is  constant  to  Fs,  where  8  is  the 
distance  one  surface  slips  on  the  other. 


248 


APPLIED   MECHANICS  FOR  ENGINEERS 


FIG.  200 


Consider  the  work  done  against  fric- 
tion on  an  axle  rotating  in  a  fixed  bear- 
ing (Fig.  200).  Here  the  point  of  ap- 
plication of  the  friction  force,  F,  changes 
in  the  body  while  the  initial  point  of  the 
force  vector  remains  fixed  in  space. 
The  rate  at  which  work  is  being  done  against  the  fric- 
tion is  then  Fv,  where  v  is  the  velocity  of  a  point  on  the 
circumference  of  the  axle,  and 
hence  the  work  done  in  turning 
through  any  angle  0  is  FrO. 

If  the  bearing  is  also  in  motion, 
as  in  Fig.  201,  then  if  tlie  bearing 
moves  forward  s  ft.  while  the  axle 
rubs  a  distance  s'  ft.  past  the  bear- 
ing, the  total  work  done  against 
friction  is  F(s  +  *')  -  Fs,  or  Fsf. 


FIG.  201 


136.  Units  of  Work.  —  The  unit  of  work  involves  the 
unit  of  space  and  the  unit  of  force.     If  the  distance  is 
measured  in  feet  and  the  force  in  pounds,  the  unit  of  work 
is  the  work  done  by  a  force  of  1  Ib.  when  the  displacement 
in  the  direction  in  which  the  force  acts  is  1  ft. 

This  unit  of  work  is  called  the  foot-pound,  designated 
by  ft.-lb. 

In  the  c.  g.  s.  system  the  erg  is  defined  as  the  work 
done  by  a  force  of  1  dyne  when  the  displacement  in  the 
direction  in  which  the  force  acts  is  1  cm. 

The  gram- centimeter  is  the  work  done  by  a  force  of  1 
gram  working  through  a  distance  of  1  cm.,  etc. 

137.  Work  of  Components  of  a  Force.  —  Let  the  constant 
force  P  (Fig.  202)  acting  on  a  body  have  the  displacement 


WORK  AND  ENERGY 


249 


D 


AB,  or  s.  Resolve  P  into  any  two  rectangular  com- 
ponents, X  and  Y.  In  the  given  movement  the  displace- 
ment of  -X"  is  AC,  or  x,  and 
of  T  is  AD,  or  y.  The  work 
done  by  P  in  the  given  dis- 
placement is  P  cos  0  •  s. 

But  P  cos  0  is  the  projec- 
tion of  P  on  J.^  and  is  there- 
fore equal  to  the  sum  of  the 
projections  of  X  and  Y  on 
AB;  i.e. 

P  cos  0  =  X  cos  a  +  Y  sin  a. 


FIG.  202 


cos  0  •  s  =  Xs  cos  (t+Ys  sin  a 


Therefore,  the  work  done  by  a  constant  force  in  any  dis- 
placement is  equivalent  to  the  work  done  by  any  rectangular 
components  of  the  force  in  the  same  displacement. 

Problem  323.  A  weight  of  20  Ib.  is  dragged  50  ft.  up  a  plane 
inclined  30°  to  the  horizon  by  a  constant  force,  P  =  25  Ib.,  acting 
at  an  angle  of  15°  to  the  plane.  There  is  a  retarding  force,  R  =  5  Ib., 
along  the  plane.  Compute  the  work  done  by,  or  against,  each  force 
acting  on  the  body.  Find  the  work  done  by  the  horizontal  and 
vertical  components  of  P  in  the  given  displacement,  and  also  the  work 
done  by  the  components  of  W  perpendicular  to  and  parallel  to  the 
given  plane. 

If  the  body  started  from  rest,  what  velocity  does  it  have  at  the  end 
of  the  50  ft.  ? 

Problem  324.  A  block,  A,  weighing  50  Ib.,  is  pulled  by  a  hori- 
zontal force  of  32  Ib.  along  a  horizontal  plane.  A  block,  B,  weighing 
20  Ib.,  rests  on  top  of  A.  If  the  coefficient  of  friction  between  A  and 


250 


APPLIED  MECHANICS  FOB  ENGINEERS 


the  plane  is  £  and  that  between  B  and  A  is  |,  will  B  slip  on  A  ?    If 
so,  how  far  would  B  slip  back  on  A  when  A  has  moved  forward  10  ft.  ? 
What  work  has  been  done  against  friction  in  this  motion  ? 

Problem  325.  In  the  preceding  problem  what  would  the  co- 
efficient of  friction  between  A  and  B  have  to  be  in  order  that  B 
would  just  not  slip?  What  then  would  be  the  total  work  done 
against  friction  when  A  has  moved  forward  10  ft.  ? 

138.  Work  of  a  Variable  Force.  —  If  a  force  acting  on  a 
definite  point  of  a  body  changes  in  magnitude  and  direction 
and  the  point  on  which  it  acts  moves  along  a  curve,  the 

,T 


FIG.  203 


work  done  in  any  displacement  from  A  to  B  (Fig.  203)  is 
defined  to  be 


Wp  = 


cos 


the  integration  being  taken  from  A  to  B,  where  (f>  is  the 
angle  at  any  point  of  the  curve  between  the  force  and  the 
tangent  to  the  curve. 

Since  ds,  dx,  and  dy  are  related  just  as  s,  z,  and  y,  in 
Art.  137,  we  may  write 


or,  if  the  curve  is  in  three  dimensions, 


WORK  AND  ENEEGT 


251 


Problem  326.  Prove  by  integration  that  the  work  done  against 
gravity  in  bringing  the, weight  Whom  the  position  A  to  the  position  B 
(Fig.  204)  along  the  quadrant  of  a  circle  is  Wr. 

Problem  327.     Show  from  the  definition,  ' 


Work  =  \  P  cos  <f>ds, 

that  when  a  weight  W  is  brought  from  any 
position  along  any  curve  to  a  position  h  ft. 
higher,  the  work  done  against  the  earth's  at- 
traction for  the  body  is  Wh. 

139.  Graphical  Representation  of  Work.  —  Work  has  been 
denned  as  the  product  of  a  force  and  a  distance.  If  the 
force  be  uniform  and  equal  to  P,  and  the  body  upon  which 
it  acts  be  moved  through  a  distance  a,  the  graphical  repre- 
sentation of  the  work  done  by  P  is  given 
by  the  area  of  a  rectangle  (Fig.  205) 
constructed  on  P  and  a  as  sides,  since 

W=Pa. 

If  the  force  P  varies  as  the  distance 
through  which  the  body  is  moved  along 

its  line  of  action,  we  may  represent  the  work  by  the  area 

of  the  triangle  as  shown  in  Fig. 

206.       Let    the    force    be    zero 

when   the   motion   begins,  and 

let  it  be  P1  when  the  distance 

passed   over   along   its   line   of 

action  is    OA.     Then  since  the 

force  varies  as  the  distance,  it 

is  equal  to  P  as  shown  in  Fig. 

206  for  any  intermediate  distance  GO.     The  total  work 


r 

a 

SPACE 

FIG.  205 

SPACE 

FIG.  206 


252 


APPLIED  MECHANICS  FOR  ENGINEERS 


done,  then,  in  moving  the  body  through  a  distance  OA  by 
the  variable  force  P,  which  varies  as  the  distance,  is  equal 

(P  OA} 

to^ — 1- — '-•     It  is  seen  that  this  is  the  same  as  the  work 

p 

done  by  the  average  force  — 1  acting  through  the  distance 

2 

OA.  The  resistance  of  a  helical  spring  varies  with  the 
elongation  or  compression.  The  same  law  of  variation 
holds  for  all  elastic  bodies. 

Another  variation  of  force  with  distance  with  which  the 
engineer  is  frequently  concerned,  is  the  case  where  the 
force  varies  inversely  as  the  distance  through  which  the  body 
is  moved.  In  this  case,  if  P  is  the  force  and  S  the  distance, 
the  relation  between  force  and  distance  may  be  expressed, 
PS '=  const. 

But  this  represents  the  equilateral  hyperbola.  This  will  be 
made  clearer  by  reference  to  the  specific  example  of  the  ex- 
pansion of  steam  in  a  steam  cylinder.  (See  Fig.  207.)  Up 

to  the  point  of 
cut-off  (7,  the 
steam  pressure 
is  the  same  as 
that  in  the  boiler 
(practically), 
and  is  constant 
while  the  piston 
moves  from  0  to 

0.   At  this  point, 
FIG.  207 

the      entering 

steam  is  cut  off  and  the  work  done  must  be  done  by  the 


WORK  AND  ENERGY 


253 


expansion  of  the  steam  now  in  the  cylinder.  According  to 
Mariotte's  Law,  the  pressure  varies  inversely  with  the 
volume  of  steam ;  but  since  the  cross  section  of  the  cylinder 
is  constant,  we  may  say  that  the  pressure  varies  inversely 
as  the  distance.  From  the  definition  of  work, 


FIG.  208 


it  follows  that  the  area  under  the  curve  represents  the  work 
done. 

The  curve  ob- 
tained .in  practice 
representing  the  re- 
lation between  the 
force  and  distance  is 
shown  in  Fig.  208. 

The    curve     after 
cut-off  is  not  a  true 
hyperbola,  and  its  area  is  determined  by  means  of  a  plani- 
meter  or  by  Simpson's  Rule. 

140.  Power.  — The  idea  of  work  is  independent  of  time. 
But  for  economical  reasons  it  is  necessary  to  take  into 
consideration  this  element  of  time.  We  must  know 
whether  certain  work  has  been  done  in  an  hour  or  ten 
hours.  For  such  information  a  unit  of  the  rate  at  which 
work  is  done  has  been  adopted.  This  unit  is  called  power. 
Power  is  the  rate  of  doing  work.  It  is  the  ratio  of  the  work 
done  to  the  time  spent  in  doing  that  work. 

The  unit  of  power  is  the  horse  power.  This  has  been 
taken  as  550  ft.-lb.  per  second,  or  33,000  ft.-lb.  per  minute. 


254  APPLIED  MECHANICS  FOR  ENGINEERS 

Originally  the  idea  of  the  rate  of  work  was  connected 
with  the  rate  at  which  a  good  draft  horse  could  do  work. 
This  value  as  used  by  Watt  was  550  ft.-lb.  per  second. 
The  horse  power  of  a  steam  engine  is  mean  effective 
pressure  times  distance  traveled  by  the  piston  per  second, 
divided  by  550. 

141.  Energy.  —  Energy  is  the  capacity  for  doing  work; 
it  is  stored-up  work.     Bodies  that  are  capable  of  doing 
work  due  to  their  position  are  said  to  possess  potential 
energy.     Bodies  that  are  capable  of  doing  work  due  to 
their  motion  are  said  to  possess  kinetic  energy.  •  An  ex- 
ample of  potential  energy  is  that  possessed  by  a  brick  on 
the  top  of  a  chimney.     If  the  brick  should  fall,  its  energy 
at  any  instant  during  the  fall  would  be  part  kinetic  and 
part  potential.     At  the  instant  of  striking  the  ground  the 
energy  is  all  kinetic.     When  the  brick  strikes  the  ground, 
work  is  done  in  deforming  the  ground  and  brick,  or  perhaps 
in  breaking  the  brick  and  in  generating  heat.     The  work 
done  by  the  brick  when  it  strikes  is  sufficient  to  use  up 
all  the  kinetic  energy  that  the  brick  had  when  it  struck. 

142.  Conservation  of  Energy.  —  The  kinetic  energy  that 
the  brick  spoken  of  in  the  last  article  had  when  it  struck 
was  used  up  in  doing  work  on  the  ground  and  air,  and  upon 
the  brick  itself.     It  was  not,  however,  destroyed,  but  was 
transferred  to  other  bodies,  or  into  heat.     Such  transference 
is  in  accord  with  the  well-known  principle  of  the  conserva- 
tion of  energy.     This  principle  may  be  stated  as  follows: 
energy  cannot  be  created  or  destroyed.    The  amount  of  energy 
in  the  universe  is  constant.     This  means  that  the  energy 


WORK  AND  ENERGY 


255 


given  up  by  one  body  or  system  of  bodies  is  transferred 
to  some  other  body  or  bodies.  It  may  be  that  the  energy 
changes  its  form  into  light,  heat,  or  electrical  energy. 

Energy  cannot  be  created  or  destroyed;  it  is,  therefore, 
evident  that  such  a  thing  as  perpetual  motion  is  impossible. 
Such  a  motion  would  involve  the  getting  of  more  energy 
from  a  system  of  bodies  than  was  put  into  them, 

143.  Relation  between  Work  and  Energy.  —  Suppose  the 
particle  of  mass  M  acted  upon  by 
any  «set  of  forces  to  move  along 
the  path  from  A  to  B  (Fig.  209). 
Resolve  the  forces  along  and  per- 
pendicular to  the  curve  at  each 
point.  Let  Ft  be  the  tangential 
component  of  the  resultant  of  all 
forces  acting  on  the  particle.  If 
at  is  the  tangential  component  of 
the  acceleration,  then 


Fia.  209 


But 


dv 


dv 


_       _  _ 

l~  dt  ~  dsdi~  V' 


(Art.  117.) 

as 

The  work  done  on  the  particle  in  going  from  A  to  B  is 

CBFtds.  (Art.  138.) 

»/A 

If  the  velocity  of  the  particle  is  VQ  at  A  and  vt  at  B^ 

.  •.  I    Ftds  =  I  1 Mvdv  =  4-  M(v*  —  v%), 
J*  c/«0 

or,  The  work  done  on  the  particle  =\M(v\-v$.          (1) 


256  APPLIED  MECHANICS  FOR  ENGINEERS 

The  kinetic  energy  of  the  particle  has  been  defined  as 
the  work  the  particle  can  do  due  to  its  velocity.  Suppose 
in  Fig.  209  Ft  is  opposite  to  the  direction  of  motion  and 
the  point  B  is  such  that  the  particle  comes  to  rest  there 
from  the  velocity  VQ  at  A.  Then,  from  the  definition, 

CB 
K.  E.  of  particle  at  A  —  I    Ftds. 

*/A 

In  this  case,  —  Ft  =  Mat. 

But  at  =  v^. 

ds 

Therefore  Ftds  =  — 

and  £BFtds  =  -C°Mvdv  =  \ 

Therefore  the  kinetic  energy  of  a  particle  with  velocity 
v  is  J  Mv2.  Equation  (1)  of  this  article  may  then  be  ex- 
pressed, The  work  done  on  a  particle  in  any  displacement 
~by  the  resultant  of  all  the  forces  acting  on  the  particle  is 
equal  to  the  gain  in  kinetic  energy  of  the  particle. 

For  any  body  having  a  motion  of  translation  only  the 
proof  applies  as  well  as  to  a  particle. 

Problem  328.  A  car  whose  weight  is  20  tons,  having  a  velocity 
of  60  mi.  per  hour,  is  brought  to  rest  by  means  of  brake  friction 
after  the  power  has  been  shut  off.  If  the  tangential  force  of  friction 
of  200  Ib.  acts  on  each  of  the  8  wheels,  how  far  will  the  car  go  before 
coming  to  rest? 

Problem  329.  Suppose  the  car  in  the  preceding  problem  to  be 
moving  at  the  rate  of  60  rni.  per  hour  when  the  power  is  shut  off, 
what  tangential  force  on  each  of  the  8  wheels  will  bring  the  car  to 
rest  in  one  half  a  mile  ? 


WORE  AND  ENERGY  257 

Problem  330.  What  kinetic  energy  per  second  is  developed  by 
a  river  200  ft.  wide  and  15  ft.  deep,  if  it  flows  at  the  rate  of  4  mi.  per 
hour,  the  weight  of  a  cubic  foot  of  water  being  62.5  lb.?  What  horse 
power  might  be  developed  by  using  all  the  water  in  the  river  ? 

Problem  331.  The  height  of  free  fall  in  Niagara  Falls  is  165  ft. 
Assuming  the  velocity  of  the  water  at  the  top  to  be  zero,  what  is  its 
velocity  at  the  foot?  The  flow  is  approximately  270,000  cu.  ft.  per 
second.  How  much  kinetic  energy  is  developed  per  second  ?  What 
H.'  P.  could  be  developed  by  using  all  the  water?  Counting  the  height 
of  the  fall,  including  rapids  above  and  below,  as  216  ft.,  what  H.  P. 
could  be  developed  by  using  all  the  water? 

NOTE.  It  is  estimated  that  the  total  horse  power  of  Niagara  Falls, 
considering  the  fall  as  216  ft.,  is  7,500,000.  The  Niagara  Falls  Power 
Company  diverts  a  part  of  the  volume  of  water  above  the  rapids  into 
their  power  plants,  where  it  passes  through  a  tunnel  into  the  river 
below  the  falls.  The  turbines  are  140  ft.  below  the  water  level,  and 
each  one  is  acted  upon  by  a  column  of  water  7  ft.  in  diameter.  The 
estimated  power  utilized  in  this  way  is  220,000  horse  power.  The 
student  should  estimate  the  horse  power  of  each  turbine,  assuming 
the  water  to  fall  from  rest  through  140  ft.  For  a  full  account  of  the 
power  at  Niagara  Falls,  the  student  is  referred  to  Proc.  Inst.  C.  E., 
Vol.  CXXIV,  p.  223. 

Problem  332.  A  body  whose  weight  is  32.2  lb.  is  pulled  up  a 
plane,  inclined  at  an  angle  of  30°  with  the  horizontal,  by  a  horizontal 
force  of  250  lb.  The  motion  is  resisted  by  a  constant  force  of  friction 
of  10  lb.  acting  along  the  plane.  If  it  starts  from  rest,  what  will  be 
its  velocity  after  it  has  gone  up  a  distance  of  100  ft.  ? 

Problem  333.  The  same  body  as  that  in  the  preceding  problem  is 
projected  down  the  plane  with  a  velocity  of  5  ft.  per  second.  How  far 
will  it  go  before  coming  to  rest  ? 

Problem  334.   Solve  Problem  240  by  work  and  energy. 
Problem  335.   Solve  Problem  242  by  work  and  energy. 

Problem  336.   Solve  Problem  243  by  work  and  energy, 
s 


258 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  337.  A  body  whose  weight  is  64.4  Ib.  falls  freely  from 
rest  from  a  height  of  5  ft.  upon  a  200-lb.  helical  spring.  Find  the 
compression  in  the  spring. 

Problem  338.  A  weight  of  500  Ib.  is  to  fall  freely  from  rest 
through  a  distance  of  6  ft.  The  kinetic  energy  is  to  be  absorbed 
by  a  helical  spring.  Specifications  require  that  the  spring  shall  not  be 
compressed  more  than  2  in.  Find  the  strength  of  the  spring  required. 

Problem  339.  Specifications  state  that  it  shall  require  32,000  Ib. 
to  compress  a  helical  spring  l£  in.  What  weight  falling  freely  from 
rest  through  a  height  of  10  ft.  will  compress  it  one  inch  ? 

Problem  340.   The  draft  rigging  of  a  freight  car  shown  in  Fig. 

210  is  provided  with  two 
helical  springs,  one  inside 
the  other.  The  outside 
spring  is  a  10,000-lb. 
spring,  and  the  inside  a 
5000-lb.  spring.  A  car 
weighing  60,000  Ib.  is  pro- 
vided with  such  a  draft 
FIG.  210  .  . 

rigging.     While  going  at 

the  rate  of  1  mi.  per  hour  it  collides  with  a  bumping  post.     How 
much  will  the  springs  be  compressed? 

Problem  341.  The  draft  rigging  in  the  preceding  problem  is 
attached  to  the  first  car  of  a  freight  train,  consisting  of  30  cars,  each 
weighing  60,000  Ib.  How  much  will  the  springs  of  the  first  car  be 
elongated  if  there  is  10  Ib.  pull  for  each  ton  of  weight  when  the  speed 
is  40  mi.  per  hour?  The  speed  is  increased  to  45  mi.  per  hour. 
How  much  will  the  springs  be  elongated  if  the  resistance  per  ton  at 
this  speed  is  12  Ib.  ? 

Problem  342.  The  Mallet  compound  locomotive  (Railway  Age, 
Aug.  9,  1907)  is  capable  of  exerting  a  draw-bar  pull  of  94,800  Ib. 
According  to  the  preceding  problem,  how  many  60,000-lb.  cars  can  be 
pulled  at  45  mi.  per  hour?  What  strength  of  spring  would  be 
necessary  for  the  first  car,  if  the  allowable  compression  is  1|  in.? 


WORK  AND  ENERGY 


259 


Problem  343.  An  automobile  going  at  the  speed  of  30  mi.  per 
hour  comes  to  the  foot  of  a  hill.  The  power  is  then  shut  off  and  the 
machine  allowed  to  "  coast "  up  hill.  If  the  slope  of  the  hill  is  1  ft. 
in  50,  how  far  up  the  hill  will  it  go,  if  friction  acting  down  the  plane 
is  .06  (7,  where  G  is  the  weight  of  the  machine? 

144.  Pile  Driver.  —  A  pile  driver  consists  essentially  of 
a  hammer  of  weight  G-  so  mounted  that  it  may  have  a  free 
fall  from  rest  upon  the  pile  (Fig.  211).  The  safe  load 
to  be  placed  upon  a  pile  after  it  has  been 
driven  is  the  problem  that  interests  the 
engineer.  This  is  usually  determined  by 
driving  the  pile  until  it  sinks  only  a  certain 
fraction  of  an  inch  under  each  blow,  then 
the  safe  load  is  a  fraction  of  the  resistance 
offered  by  the  earth  to  these  last  blows. 
The  resistance  is  usually  small  when  the  pile 
begins  to  penetrate  the  earth,  but  increases 
as  penetration  proceeds,  until  finally,  due 
to  the  last  blows,  it  is  nearly  constant. 
If  WQ  regard  the  hammer  G-  as  a  freely 
falling  body,  and  consider  the  hammer 
and  pile  as  rigid  bodies,  and  further 
assume,  as  is  usually  done,  that  R  for  the 
last  few  blows  is  constant,  we  may  write 
the  work-energy  equation,  ^IQ.  211 


since  the  final  kinetic  energy  is  zero  and  the  weight  of  the 
hammer  as  a  working  force  is  negligible.  The  distance  81 
is  the  amount  of  penetration  of  the  pile  for  the  blow  in 


260  APPLIED  MECHANICS  FOR  ENGINEERS 

question.     But  v$  =  2  gh,  so  that 
l  Mv*  =  ah. 

We  have  then  as  the  value  for  the  supporting  power  of 
a  pile, 


A  safe  value,  R(  from  ^  to  -J  of  R,  is  taken  as  the  safe  sup- 
porting power  of  piles.  The  factor  of  safety  and  the  value 
of  81  for  the  last  blow  are  usually  matters  of  specification 
in  any  particular  work.  This  is  the  formula  given  by 
Weisbach  and  Molesworth.  Other  authorities  give 
formulas  as  follows  : 

Trau twine,    R  =  60  G-\/h,  if  81  is  small, 
and  R  =  5  ^^ ; 

Wellington,  R  =  —    ^— ,  where  h  is  in  feet  and  sl  in  inches  ; 
£]_  H~  -L 

Me  Alpine,     R  =  80  [  G-  +  (.  228  VX  - 1)  2240]  ; 

Goodrich,      R  = 

8% 

For  other  formulas  and  a  general  discussion  of  the  sub- 
ject of  the  bearing  power  of  piles,  the  reader  is  referred  to 
Trans.  Am.  Soc.  Civ.  Eng.,  Vol.  48,  p.  180. 

The  great  number  of  formulas  for  the  supporting  power 
of  piles  is  due  to  the  various  assumptions  that  are  made 
in  deriving  them.  In  deriving  the  Molesworth  formula^, 
the  hammer  and  pile  were  considered  as  rigid  bodies.  It 
will  be  seen  that  the  hammer  and  pile  are  both  elastic 


WORK  AND  ENERGY  261 

bodies,  both  are  compressed  by  the  blow;  there  is  friction 
of  the  hammer  with  the  guides,  and  the  cable  attached  to 
the  hammer  runs  back  over  a  hoisting  drum.  There  is, 
in  most  cases,  a  loss  of  energy  due  to  brooming  of  the  head 
of  the  pile.  This  broomed  portion  must  be  cut  off  before 
noting  the  penetration  due  to  the  last  few  blows. 

Results  of  tests  have  also  been  taken  into  consideration 
and  have  modified  the  formulae.  The  Wellington  formula 
differs  from  the  Molesworth  formula  only  in  the  denomi- 
nator, where  s1  +  1  is  used  instead  of  sr  This  has  been 
done  to  guard  against  the  very  large  values  of  R  given 
when  s1  is  very  small. 

Engineers  have  come  to  believe  that  it  will  be  extremely 
difficult  to  get  a  general  formula  that  will  give  very  exact 
information  as  to  the  bearing  power  of  piles,  since  soil  con- 
ditions are  so  varied.  The  more  simple  formula  with  a 
proper  factor  of  safety  are  used. 

As  an  illustration  of  the  use  of  the  formula,  let  us  con- 
sider the  problem  of  providing  a  pile  to  support  75  tons. 
If  the  weight  of  the  hammer  is  3000  lb.,  and  the  height  of 
fall  15  ft.,  the  pile  will  be  considered  down  when 

G-h     3000  x  15       o  f  .       o  fi  - 

; 


Using  a  factor  of  safety  of  6,  we  have  sl  =  .6  in. 

Problem  344.  Compute  the  value  of  sl  for  the  pile  in  the  above 
illustration  by  using  the  various  formulse  given  in  this  article.  Com- 
pare the  results. 

Problem  345.  A  pile  is  driven  by  a  4000-lb.  hammer  falling  freely 
20  ft.  What  will  be  the  safe  load  that  the  pile  will  carry  if  at  the  last 


262 


APPLIED  MECHANICS  FOR  ENGINEERS 


blow  the  amount  of  penetration  was  f-  in.?  Use  a  factor  of  safety  of  4. 
Compute  by  the  Molesworth  and  the  Wellington  formulae,  and 
compare. 

Problem  346.  A  pile  was  driven  by  a  steam  hammer.  The  last 
twenty  blows  showed  a  penetration  of  one  inch.  If  two  blows  of  the 
steam  hammer  cause  the  same  penetration  as  one  blow  from  a  2000-lb. 
hammer  falling  20  ft.,  what  weight  in  tons  will  the  pile  support  ? 
Assume  the  penetration  for  each  of  the  last  few  blows  the  same. 

145.  Steam  Hammer. —  The  steam  hammer  consists  es- 
sentially of  a  steam  cylinder  mounted  vertically  and  having 
a  weight  or  hammer  attached  to  one 
end  of  the  piston  rod.  Let  AB 
(Fig.  212)  be  the  steam  cylinder, 
D  the  piston,  and  H  the  anvil,  upon 
which  a  piece  of  metal  is  shown  un- 
der the  hammer  Gr.  The  steam 
pressure  in  the  cylinder  is  constant 
and  equal  to  P,  while  the  piston 
passes  over  a  distance  a  to  cut-off, 
and  varies  inversely  as  the  volume 
over  the  remaining  distance  b.  G-, 
the  weight  of  the  hammer  and 
piston,  is  also  a  working  force. 
a  The  resistance  of  the 


]    metal  varies  during  any 


blow  with  the  amount  of  compres- 
sion. It  is  zero  just  as  the  hammer 
touches  the  metal,  and  increases  up 
to  a  maximum  when  the  compres- 
sion is  greatest.  Let  R'  be  the  average  resistance  of  the 


FIG.  212 


WORK  AND  ENERGY  263 

metal,  and  E1  the  exhaust  pressure.  Then  the  work-energy 
equation  for  the  hammer  when  it  has  moved  a  distance  * 
becomes 


2          2 
Since  P'  varies  inversely  as  the  volume  of  the  cylinder, 

.,       r>f      const.      c 
we  may  write,  P  =  --  -  —  =  -  • 
7rr2s       s 

Then  the  work-energy  equation  gives 

^  +  ElS  =  Pa  +  as  +  c 

" 


The  term  =°  is  zero,  since  the  motion  has  been  consid- 

ered from  rest  at  the  top  of  the  cylinder  to  a  distance  s. 
The  quantity  c  may  be  computed  by  reading  from  the  in- 
dicator card  the  value  of  P'  at  8,  or  computed  from  c  =  Pa. 
It  will  be  seen  that  8  has  been  taken  greater  than  a  ;  that  is, 
the  piston  is  beyond  the  point  of  cut-off. 

When  the  hammer  finally  comes  to  the  face  of  the  metal, 
the  work-energy  equation  may  be  written 


where  the  distance  h'  represents  the  value  of  8  when  the 
hammer  just  touches  the  metal,  and  v^  the  corresponding 
value  of  v.  This  equation  gives  the  kinetic  energy  of  the 
hammer  when  it  strikes  the  metal.  The  work-energy 


264  APPLIED  MECHANICS  FOR  ENGINEERS 

equation  for  the  hammer  during  the  compression  of  the 
piece  may  now  be  written 


where  d  is  the  amount  of  compression  of  the  metal  due  to 
the  blow.  This  is  shown  by  the  small  figure  to  the  right, 
where  the  piece  of  metal  has  been  drawn  to  a  somewhat 
larger  scale. 

After  the  hammer  strikes  the  metal,  the  steam  pressure 
and  the  weight  of  the  hammer  as  working  forces,  and  the 
exhaust  pressure  as  a  resisting  force,  have  been  neglected. 
The  work  done  by  these  pressures  is  small,  since  the  dis- 
tance d  is  small.  Approximately,  then,  the  work  done  on 
the  metal  equals  the  kinetic  energy  at  the  time  of  first 
contact. 

Instead  of  using  the  value  P'  =  -,  and  computing  the 

integral  j  P'ds  as  indicated  in  the  formulse,  values  of  Pf 

and  s  might  be  read  from  the  indicator  diagram  (Art.  139) 
and  added  by  means  of  Simpson's  formula  (Art.  40). 

As  an  illustration  of  the  foregoing,  let  us  suppose  the 
steam  cylinder  25  in.  long  and  14  in.  in  diameter;  the 
steam  pressure  P  =  18,000  Ib.  ;  the  exhaust  pressure 
^  =  2300  Ib.;  a  =  7.2  in.;  d=\  in.;  #=644  Ib.; 
c=  10,800;  hf  =24  in.  Substituting  in  the  work-energy 
equation,  we  have  for  the  kinetic  energy  of  the  hammer  at 
the  time  of  striking  the  iron, 


=  20,490  ft.  -Ib. 

2 


WORK  AND  ENERGY  265 

This  gives  a  value  for  v1  =  45.3  ft.  per  second  as  compared 
with  11.3  ft.  per  second  for  the  same  weight  freely  falling 
through  the  same  distance. 

Investigating  now  the  resistance  of  the  metal,  we  have, 
under  the  assumption  already  made, 

R'd=  20,490  ft.-lb., 
so  that  R'  =  983,500  Ib. 

In  the  above  discussion  we  have  neglected  the  compres- 
sion of  the  anvil  and  hammer  due  to  the  blow,  and  also 
the  friction  of  the  piston. 

Problem  347.  Find  the  kinetic  energy  of  the  hammer  when 
h'  =  18  in.  Find  also  v  and  R',  using  the  same  value  of  d. 

Problem  348.  A  steam  hammer  exactly  similar  to  the  one  given 
in  the  illustration  above  is  used  with  a  different  steam  pressure.  It  is 
only  necessary  for  the  work  for  which  it  is  intended,  that  the  kinetic 
energy  of  the  hammer  for  a  stroke  of  2  ft.  be  10,000  ft.-lb. 

If  a  =  7.2  in.,  Rl  =  1800  Ib.,  find  what  steam  pressure  P  is 
necessary?  (Notice  that  c  has  a  different  value  from  that  given 
above.) 

Problem  349.  Compute  the  kinetic  energy  and  velocity  of  the 
hammer  in  the  illustration  (G  ==  644  Ib.)  when  the  piston  has  moved 
the  full  length  of  the  cylinder  (h  =  25  in.).  Assume  that  there  is 
nothing  on  the  anvil. 

Problem  350.  What  value  of  h'  in  the  above  problem  would  give 
the  hammer  the  same  velocity  as  it  would  have  if  it  fell  freely  from 
rest  through  the  height  h  ?  Compute  the  kinetic  energy  for  this 
velocity. 

Problem  351.  In  the  illustration  given  above,  what  would  be 
the  value  of  R'  if  the  steam  pressure  and  G  be  included  as  working 
forces,  and  Rl  as  a  resisting  force,  during  the  compression  of  the 
piece  ? 


266 


APPLIED  MECHANICS  FOE  ENGINEERS 


Problem  352.  In  the  illustration  given  above,  suppose  that,  in 
addition  to  the  compression  of  the  piece,  \  in.,  the  anvil  is  compressed 
.02  in.  Find  the  value  of  R' '. 

146.  Rotation  of  a  Body  about  a  Fixed  Axis.  —  For  a  body 
rotating  about  a  fixed  axis,  if  the  angle  turned  through  is 
0,  the  angular  velocity  at  any  instant  is  defined  as 


FIG.  213 


and 


and  the  angular  acceleration  as 

0=^, 

which  may  also  be  written 

_  d<adQ_    do> 
~dQdt~    dQ* 


For  a  particle  describing  a  circle  of  radius  r,  if  v  is  the 
velocity  of  the  particle  at  any  time  and  at  the  tangential 
component  of  the  acceleration,  then  (Fig.  213), 

ds        dd 

v  —  —  =  r  —  =  rut, 
dt        dt 


and 


dv        dco 

a>t  =  —  =  r  — 

dt         dt 


r*. 


147.  Work  Done  by  Impressed  Forces  on  a  Rotating  Body.  — 
In  Fig.  214  is  represented  a  thin  slab  of  a  body  rotating 
about  a  fixed  axis  AB,  made  by  planes  perpendicular  to 
AB.  Let  the  body  be  acted  upon  by  any  set  of  external 
or  impressed  forces,  of  which  Pv  P2,  •••  are  the  components 


WORK  AND  ENERGY  267 

in  planes  perpendicular  to  AB,  the  other  components,  not 
shown,  being  parallel  to  AB. 

Let  dM  be  the  mass  of  an  element  of  the  body  distant  r 
from  AB.  At  any  instant 
dM  has  an  acceleration  due 
to  the  forces  acting  on  it. 
All  the  forces  acting  on  dM 
may  be  resolved  into  three 
components  respectively  par- 
allel to  AB,  along  the  radius, 
dN,  and  along  the  tangent, 
dT.  The  forces  dN  and  dT 
are  called  the  effective  forces.  FlG 

They   are   the   forces  which 

acting  on  dM  separated  from  the  body  would  give  it  the 
same  motion  that  it  has  as  part  of  the  body. 
.     If  at  is  the  tangential  acceleration  of  dM  and  a  is  the 
angular  acceleration  of  the  body,  then 


The  forces  dN  and  dT  are  the  components  of  the  result- 
ant of  all  forces  acting  on  dM.  The  forces  are  made  up 
of  the  action  of  forces  exerted  by  adjacent  particles  on  dM 
and  any  external  forces  that  act  011  dM.  The  forces  ex- 
erted by  the  particles  on  each  other  are  assumed  to  be  in 
the  lines  joining  the  particles,  and  to  be  equal  and  opposite, 
according  to  Newton's  laws.  Hence  in  any  summation  of 
all  the  forces  d  T  and  dN^  acting  on  all  the  particles,  the 
reactions  of  the  particles  on  each  other  would  annul  and 
leave  only  the  summation  of  the  impressed  forces.  There- 


268 


APPLIED  MECHANICS  FOB   ENGINEERS 


fore  the  sum  of  the  moments  of  the  effective  forces  about 
the  line  AB  is  equal  to  the  sum  of  the  moments  of  the  im- 
pressed forces  about  that  line  ;  i.e. 

j  rdT  =  2 (mom  of  impressed  forces)  =  2Pa. 

But          dT=radM. 

.-.  ^Pa  =   Cr*adM=  afr*dM, 
or  S  Pa  =  a  I, 

where  I  is  the  moment  of  inertia  of  the  body  about  the 
axis  of  rotation. 

The  name  moment  of  inertia  is  suggested  for  J  r2d M 

since  it  is  seen  to  be  equivalent  to  the  moment  of  a  force 
which  would  produce  unit  angular  acceleration  of  the 

body  about  the  given  axis, 

\.  against  the  inertia  of  the- 

^^  .P       body. 

The  work  done  by  any 
force  P  during  the  rota- 
tion through  the  angle  dO  is 
Pcos  $d8  (Fig.  215).  But 


ds  =  rdO,     and    cos  <£  =  - , 


FIG.  215 


and  hence  the  expression 
for  the  work  may  be  written  J  PadB,  and  the  total  work 
done  by  all  the  impressed  forces  for  any  rotation  becomes 


WORK  AND  ENERGY 


269 


It  was  shown  above  that 
done  may  be  written 


=  «J,  and  hence  the  work 


1  1  add,  or  since  a  =  co  -^, 
^  du 

.  -.Work  done  =  I  C^codco  =  \  J(«J  _  «|j), 
«/«0 

where  o>0  and  o^  are  the  initial  and  final  angular  velocities 
of  the  body  for  the  given  displacement. 

It  follows  that  the  work  that  the  rotating  body,  with 
angular  velocity  a>0,  could  do  against  resisting  forces  before 
coming  to  rest  is  J/«$.  That  is,  the  kinetic  energy  of  a 
body  rotating  about  a  fixed  axis  with  angular  velocity  o>  is 
J  Jo)2,  where  /is  the  moment  of  inertia  of  the  body  about 
the  axis  of  rotation. 

As  an  illustration,  let  us  consider 
the  case  of  two  weights  (See  Fig. 
216),  tfj  =  20  Ib.  and  #2  =  10  lb., 
suspended  from  drums  rigidly  at- 
tached to  each  other  and  of  radii  3 
ft.  and  2  ft.  respectively.  Let  the 
weight  of  the  two  drums  and  shaft 
be  644  lb.,  and  the  radius  of  gyration 
2  ft.  The  radius  of  the  axle  is  one 
inch  and  the  axle  friction  30  lb.  The 
friction  acts  tangentially  to  the  axle. 

Assuming  that  the  initial  angular     • 

velocity  o>0  is  one  radian  per  second,  and  the  final  angular 
velocity  18  radians  per  second,  how  many  revolutions  will 
the  drums  make? 


270  APPLIED  MECHANICS  FOR  ENGINEERS 

The  external  forces  acting  on  the  drums  are  the  ten- 
sions, 2\  and  T<p  in  the  cords  attached  to  the  weights  Gr1 
and  6r2  respectively,  the  reaction  of  the  axle,  and  the  force 
of  friction.  Acting  on  the  weight  Grl  are  the  forces  Gr1 
and  TI  and  on  6r2  the  forces  6r2  and  T2.  The  work-energy 
equations  for  these  three  bodies  are  respectively, 

2  irr^nTi  -  2  7rr^nT2  -  2  irrBn  .  30  =  }  80(182  -  I2), 

a,  -  2\)  =  1^1  [(18  ri)a  -  rf], 

t/ 


where  rx  is  the  radius  of  the  large  drum,  r2  that  of  the 
small  drum,  and  r3  that  of  the  axle. 

Eliminating  T^  and  T2  by  adding  the  three  equations, 
2  TTT-^J  -  2  7rr27ia2  -  2  -jrr^n  .  30 

=  V8o  +^i 

2V  g          g 

Substituting  the  known  values,  we  obtain  n  =  59.5. 

Problem  353.     In  the  above  illustration,  what  are  the  velocities  of 

(?!  and  G2  when  w  has  its 
initial  and  final  values  ?  In 
what  time  do  the  drums 
make  the  59.5  revolutions? 

Problem  354.  The  drum 
in  Fig.  217  is  solid  and  has 
a  radius  r  and  a  thickness  k. 

•  Initially,  it  is  rotating,  inak- 
ing  wo  radians  per  second, 

but  it  is  brought  to  rest  bv 
FIG.  217 

the  action  of  a  brake.     The 

brake  is  applied  from  below  by  a  force  P  acting  at  the  end  of  the  beam. 


WOES:  AND  ENERGY  271 

The  force  of  friction  between  the  drum  and  brake  is  —  ,  where  P1  is 

4 

the  normal  pressure  everted  by  the  beam  on  the  drum.  The  radius 
of  the  axle  is  rv  and  the  axle  friction  (.05)  P'1,  where  P"  is  the  pres- 
sure of  the  axle  on  the  bearing.  Required  the  work  -energy  equation. 
Since  the  drum  comes  to  rest,  the  final  kinetic  energy  is  zero,  so 
that 


2  +       2  TTTH  +  (.05)  P"2  Tr/yi  =  0. 

There  are  no  working  forces,  so  we  find  the  equation  reducing  to  the 
form  :  the  initial  kinetic  energy  equals  the  work  of  resistance.  The 
normal  pressure  exerted  by  the  beam  on  the  drum  may  be  found  by 
taking  moments  about  the  hinge  of  the  beam.  Then 

p,  =  a  +  bp 

b 

The  number  of  revolutions  turned  through  in  coming  to  rest  is 
designated  by  n.     The  equation  then  becomes 

+    Q5    p,,2       nm 


Problem  355.  Suppose  the  drum  in  the  preceding  problem  to 
be  3  ft.  in  diameter,  1£  in.  thick,  and  made  of  cast  iron.  It  is  mak- 
ing 4  revolutions  per  second  when  the  force  P  =  100  Ib.  is  applied  to 
the  beam.  The  length  of  the  drum  is  6  ft.,  and  the  rim  weighs  twice 
as  much  as  the  spokes  and  hub.  If  k  =  1.25  ft.,  a  =  b  =  4  ft.,  and 
rx  =  1  in.,  find  the  number  of  revolutions  that  the  drum  will  make 
before  coming  to  rest.  Assume  the  friction  of  the  brake  on  the  drum 
to  be  £  the  normal  pressure,  and  the  friction  of  the  axle  (.05)  P". 

Problem  356.  The  drum  in  the  preceding  problem  is  making 
3  revolutions  per  second.  What  force  will  be  required  to  bring  it 
to  rest  in  100  revolutions? 

Problem  357.  If  the  brake  in  Problem  355  is  above  instead  of 
below  the  drum,  how  will  the  results  in  Problems  355  and  356  be 
changed  ? 


272 


APPLIED  MECHANICS  FOR  ENGINEERS 


FIG.  218 


Problem  358.  A  square  prism  as  shown  in  Fig.  218  is  mounted 
so  as  to  rotate  due  to  the  weight  G.  The  elastic  cord  runs  over  the 

pulley  B  and  meets  the  square 
at  P't  but  is  connected  to  the 
square  at  P.  The  mechanism 
is  such  that  motion  begins  when 
P  is  in  the  position  shown,  and 
ceases  when  the  prism  has  made 
a  quarter  turn ;  that  is,  when  P 
reaches  P'.  The  diameter  of 
the  journal  is  2  in.,  and  the 
weight  on  the  same  is  600  Ib. 
The  force  of  friction  on  the  journal  is  60  Ib.,  and  on  the  pulley  at  B 
equivalent  to  10  Ib.  acting  at  the  rim  of  the  pulley.  Find  the  tension  in 
the  cord  when  P  reaches  P'.  The  cord  is  elastic,  and  is  made  of  such 
material  that  it  elongates,  due  to  a  pull  of  100  Ib.,  .02  in.  in  each  inch  of 
length.  What  is  the  elongation  per  inch  due  to  the  fall  of  G  as  stated  ? 

148.  Brake-shoe  Testing  Machine.  —  The  brake-shoe  testing 
machine  owned  by  the  Master  Car  Builders'  Association  has 
been  established  at  Purdue  University.  It  consists  of  a 
heavy  flywheel  attached  to  the  same  axle  as  the  car  wheel. 
These  are  connected  with  the  engine,  and  may  be  given  any 
desired  rotation.  When  this  has  been  obtained,  they  may 
be  disconnected  and  allowed  to  rotate.  The  dimensions  and 
weight  of  the  parts  are  known  so  that  the  kinetic  energy  of 
the  flywheel  and  rotating  parts  may  be  computed  by  noting 
the  angular  velocity.  When  the  desired  velocity  has  been 
attained,  the  brake  shoe  is  brought  down  on  the  car  wheel. 
The  required  normal  pressure  on  the  shoe  at  A  (See  Fig. 
219)  is  obtained  by  applying  suitable  weights  at  B.  The 
system  of  levers  is  such  that  one  pound  at  B  gives  a  nor- 
mal pressure  of  24  Ib.  on  the  brake  shoe.  The  weight  of 


WORK  AND  ENERGY 


273 


the  levers  themselves  gives  a  normal  pressure  of  1233  Ib. 
Provision  is  also  made  for  measuring  the  tangential  pull  of 
the  brake  friction  ;  this,  however,  is  not  shown  in  the  figure. 


FIG.  219 

The  weight  of  the  flywheel,  car  wheel,  and  shaft,  and 
all  rotating  parts  is  12,600  Ib.,  and  the  radius  of  gyration 
is  V2.16.  The  weight  of  12,600  Ib.  is  supposed  to  be  the 
greatest  weight  that  any  bearing  in  passenger  or  freight 
service  will  be  called  upon  to  carry.  The  diameter  of  the 
flywheel  is  48  in.,  its  thickness  30  in.,  diameter  of  shaft 
7  in.,  and  the  diameter  of  the  car  wheel  is  33  in.  The 
brake-shoe  friction  is  J  the  normal  pressure  of  the  brake 
shoe  on  the  wheel,  and  the  journal  friction  may  be 
assumed  as  (.002)  of  the  pressure  of  the  axle  on  the 
bearing.  The  work-energy  equation  for  the  rotating 
parts  after  being  disconnected  from  the  engine  becomes 


+  (1233  +  12,600  +  24  <?)(.002)2  -n-  -L  n 


274 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  359.  The  speed  is  such  as  to  correspond  to  a  speed  of 
train  of  a  mile  a  minute  when  brakes  are  applied.  What  must  be 
the  weight  G  so  that  a  stop  may  be  made  in  a  thousand  feet  ?  What 
is  the  corresponding  normal  pressure  on  the  brake  shoe? 

Problem  360.  If  the  speed  corresponds  to  the  speed  of  a  train 
of  100  mi.  per  hour,  what  weight  G  would  be  necessary  to  reduce  the 
speed  to  60  mi.  per  hour  in  one  mile  ?  What  is  the  normal  pressure 
on  the  brake  shoe  necessary? 

Problem  361.  If  the  velocity  corresponds  to  a  train  velocity  of 
60  mi.  per  hour,  and  the  apparatus  is  brought  to  rest  in  220  revolu- 
tions, the  weight  G  is  100  Ib.  Find  the  tangential  force  of  friction 
acting  on  the  face  of  the  wheel.  What  relation  does  this  bear  to  the 
normal  brake-shoe  pressure? 

NOTE.  In  the  preceding  problems,  the  ratio  (the  coefficient  of  fric- 
tion, see  Art.  135)  has  been  taken  as  \.  One  of  the  important  uses 
of  this  testing  machine  is  to  determine  the  coefficient  of  friction  for 
different  types  of  brake  shoes.  Experiment  shows  that  it  varies 
generally  from  \  to  £,  sometimes  going  as  high  as  T3Q. 

149.  Kinetic  Energy  of  a  Body  having  Plane  Motion. —  Sup- 
pose a  body  to  have  plane  motion.     Let  the  angular  veloc- 
ity of  the  body  at  a  given 
instant  be  co  and  the  ve- 
locity   of    the   center    of 
gravity  of  the  body  be  vr 
Choose  coordinate  axes 
so  that   the   origin   coin- 
cides at  the  given  instant 
with  the  center  of  gravity 
and  the  #-axis   coincides 
FlG>22°  with     the     direction     of 

motion  of  the  center  of  gravity,  the  x-  and  2-axes  being  in 
the  plane  of  motion  of  the  body  (Fig.  220). 


WORK  AND  ENERGY  275 

Let  dM  be  any  element  of  mass  of  the  body,  distant  r 
from  the  ^-axis.  The  velocity  of  dMis  then  composed  of 
the  velocity  of  any  point  on  the  #-axis  and  the  velocity  of 
dM  relative  to  the  y-axis.  Hence,  v,  the  velocity  of  dM, 
is  given  by 

2  V-Q)  COS  < 


where  <£  is  the  angle  which  r  makes  with  a  line  parallel  to 
the  z-axis.     But  r  cos  0  =  2,  so  that 


The  kinetic  energy  of  the  whole  body  is  the  sum  of  the 
kinetic  energy  of  its  particles,  or 


K.  E.  of  Body  =     j(vf  + 

=  1  v\  CdM  + 

Since  the  a?y-plane  passes  through  the  center  of  gravity, 
dM=  0.  (Art.  33.) 


Therefore 

K.  E.  of  Body  =  |  Mv\  + 1 1«2, 

where  Jf  is  the  mass  of  the  body  and  /  its  moment  of 
inertia  about  a  gravity  axis  perpendicular  to  the  plane  of 
motion. 

This  formula  may  be  expressed  in  words  as  follows  : 
The  kinetic  energy  of  a  body  having  plane  motion  is  equal 
to  the  kinetic  energy  the  whole  mass  would  have  if  concen- 
trated at  the  center  of  gravity,  with  the  velocity  of  the  center 
of  gravity,  plus  the  kinetic  energy  of  rotation  that  the  body 


276  APPLIED  MECHANICS  FOR   ENGINEERS 

would  have  if  the  gravity  axis  were  at  rest  and  the  body 
rotating  about  it  with  the  same  angular  velocity. 

150.  Work  and  Kinetic  Energy  in  Plane  Motion.  —  If  im- 
pressed forces  act  on  a  body  having  plane  motion,  the 
work  done  by  these  forces  in  any  displacement  equals  the 
change  in  kinetic  energy  of  the  body  in  that  dis- 
placement. 

Proof:  The  total  gain  in  the  kinetic  energy  of  the  body 
is  the  sum  of  the  increments  in  kinetic  energy  of  all  the 
particles  of  the  body,  which,  by  Art.  143,  is  the  total 
work  done  on  the  particles  by  all  the  forces  acting  on 
them.  In  finding  the  total  work  done  on  the  particles 
the  work  done  by  the  forces  that  the  particles  exert  on 
each  other  adds  up  to  zero  and  there  is  left  only  the  work 
done  by  the  impressed  forces.  For  two  particles  of  the 

body,  m1  and  m^  remain  the  same 
distance  apart  and  exert  on  each 
other  forces  that  are  equal  and 
opposite  and  in  the  line  joining 
the  particles.  During  any  dis- 
placement w2  has  a  motion  made 
up  of  the  motion  of  m,  and  a 

FIG.  221 

motion  about  m^  as  a  center  (Fig. 

221).  For  that  component  of  its  motion  parallel  to  the 
motion  of  ml  the  forces  exerted  by  the  particles  on  each 
other  are  equal  and  opposite  and  have  the  same  displace- 
ment, and  hence  the  work  done  by  one  force  is  just  equal 
to  the  work  done  against  the  other.  For  the  motion  of 
rotation  of  w2  about  m±  the  force  acting  on  w2  is  perpen- 


WORK  AND  ENERGY  277 

dicular  to  the  direction  of  displacement,  and  hence  no  work 
is  done.  Hence  the  forces  exerted  by  the  particles  on  each 
other  do  no  work,  and  we  may  write  for  any  plane  motion, 

Work  done  by  impressed  forces  =  change  in  kinetic  energy  of  body. 

As  an  illustration,  consider  a  body  of  circular  section, 
as  a  hoop,  cylinder,  or  sphere,  with  center  of  gravity  at 
the  center  of  the  circular  section,  rolling  without  slipping 
down  an  inclined  plane  (Fig.  222).  The  impressed 
forces  acting  on  the  body  are 
its  weight,  Gr,  the  normal 
reaction  of  the  plane,  JVJ  and 
a  retarding  friction  force,  F, 
along  the  plane.  At  any  in- 
stant the  point  of  the  body  FIG.  222 

at  which  F  and  N  are  ap- 

plied is  the  instantaneous  center  of  rotation  (compare 
Art.  131)  and  hence  its  velocity  is  zero.  Therefore  the 
rate  at  which  the  forces  F  and  N  are  doing  work  are 
throughout  the  motion  zero.  (See  Art.  134.)  Hence  no 
work  is  done  by  or  against  these  forces  in  the  displace- 
ment due  to  rolling.  The  total  work  done  in  the  descent 
is  therefore  Grh. 

If  o>0  and  VQ  are  the  angular  velocity  of  the  body  and 
the  linear  velocity  of  its  center  of  gravity  respectively  at 
the  top  of  the  plane  and  co  and  v  the  corresponding  values 
at  the  foot,  then  the  work-energy  equation  is 


If  s  is  the  distance  passed  through  by  the  center  and  0 


278  APPLIED  MECHANICS  FOR  ENGINEERS 

the  angle  turned  through  in  the  same  time, 


Therefore  l  =  r 

dt       dt 

or  v  =  ro). 

Substituting  co  =  -,  COQ  =  \  1=  —  k2,  M=  -, 

r  r  g  g 

the  work-energy  equation  becomes 


Problem  362.  Prove  that  all  solid  spheres  will  roll  down  the 
inclined  plane  at  the  same  rate.  Find  the  velocity  at  the  foot  of  the 
plane. 

Problem  363.  A  uniform  sphere,  a  uniform  disk,  and  a  hoop, 
starting  at  the  top  of  an  inclined  plane,  roll  from  rest  to  the  foot. 
Find  the  velocity  of  each  on  reaching  the  foot.  In  what  order  do 
they  arrive  ? 

Problem  364.  Which  will  roll  faster  down  an  inclined  plane,  a 
hollow  sphere  with  diameter  of  the  hollow  one  half  that  of  the 
sphere,  or  a  solid  uniform  disk  ? 

151.  Kinetic  Energy  of  Rolling  Bodies.  —  It  is  convenient 
to  express  the  kinetic  energy  or  combined  rotation  and 
translation  of  such  bodies  as  rolling  wheels  in  a  different 
form  from  that  given  in  the  preceding  article.  There  is 
some  mass  M^  that  will  have  the  same  kinetic  energy 
when  translated  with  a  velocity  v1  as  the  kinetic  energy 


WORK  AND  ENERGY  279 

of  translation  plus  the  kinetic  energy  of  rotation  of  the 
body  of  mass  M\  that  is, 

Mrf  __  Mv\ 


2  2    "  2 

For  a  wheel  rolling  on  a  straight  track  cor  =  vv  where  r  is 
the  radius. 

Then  Ml 

This  has  been  called  the  equivalent  mass. 

For  example,  for  a  rolling  disk,  since  /=  J  Mr2,  M1  =  %M. 

Problem  365.  A  sphere  rolls  without  slipping  down  an  inclined 
plane.  Show  that  its  kinetic  energy  is  the  same  at  any  instant  as  that 
of  a  sphere  whose  mass  is  f  larger  translated  with  a  velocity  equal  to 
the  velocity  of  the  center  of  gravity  of  the  rolling  sphere. 

Problem  366.  The  sphere  in  the  preceding  problem  is  made  of 
steel,  12  in.  in  diameter,  and  the  inclination  of  the  plane  is  30°.  If 
vQ  =  10  ft.  per  second,  what  will  be  the  velocity  10  ft.  down  the  plane  ? 

152.  Work-energy  Relation  for  Any  Motion.  —  The  rela- 
tion between  work  and  energy  for  the  motions  considered 
in  this  chapter  holds  for  more  complicated  motions  and 
for  motions  in  general.  The  limits  of  the  present  work 
will  not  admit  the  proof  of  the  general  theorem.  It  may 
be  said,  however,  that  for  any  motion  the  work  done  by 
the  working  forces  equals  the  work  done  by  the  resisting 
forces  plus  the  change  in  kinetic  energy.  In  the  case  of 
the  motion  of  a  complicated  machine,  the  work  done  by 
the  working  forces  equals  the  work  done  against  the  resist- 
ances plus  the  gain  in  kinetic  energy  of  the  various  parts 
of  the  machine. 


280  APPLIED  MECHANICS  FOB   ENGINEERS 

153.  Work  Done  when  Motion  is  Uniform.  —  When  the 
motion  is  uniform,  the  change  in  kinetic  energy  is  zero, 
and  the  work-energy  equation  reduces  to  the  form  :  work 
done  equals  the  work  done  against  the  resistance  overcome. 

As  an  illustration,  let  us  consider  the  case  of  a  loco- 
motive moving  at  uniform  speed  and  represented  in  Fig. 
223.  Suppose  P  the  mean  effective  steam  pressure  (See 
Art.  139),  F  the  friction  of  the  piston,  F'  the  friction  of 
the  crosshead,  F"  the  journal  friction,  F'"  the  crank-pin 


FIG.  223 

friction,  T  the  friction  on  the  rail,  R  the  draw-bar  resist- 
ance, R1  the  horizontal  component  of  pressure  of  the 
drivers'  axle  on  the  frame,  r,  rv  and  r2  the  radii  of  the 
crank-pin  circle,  the  driver  axle,  and  the  crank-pin  re- 
spectively, G-  the  weight  of  the  locomotive,  and  N1  and  N 
the  normal  reactions  of  the  rails  on  the  wheels.  Consider 
both  sides  of  the  locomotive  and  write  the  work-energy 
equation  for  a  distance  s,  equal  to  a  half  turn  of  the  driver 
(from  dead  center  A  to  dead  center  .#),  that  the  locomo- 
tive travels. 

Considering  the  work  done  on  the  frame,  counting  both 
cylinders,  we  have 

R'Tra  +  2(JF  +  F')ira  +  2  F"ira  =  (2 P  +  R)ira.   (1) 


WORK  AND  ENERGY  281 

Considering  the  work  done  on  the  rotating  and  oscillat- 
ing parts, 

2  P(ira  +  2  r)  =*  J2'7ra  +  2(P  +  I")(ira  +  2  r) 

+  2  ^"(™+^i)  +  2  ^"Wa.  (2) 

Adding  these  equations  there  results 

IPr  =  JRira  +(.F+  F')lr  +  2  JF"im  +  2.F'"irr2.         (3) 

If  we  neglect  friction,  this  equation  becomes 


or 


Here  P  is  the  mean  effective  pressure  in  one  cylinder. 

This  is  the  formula  usually  given  for  the  tractive  power 
of  a  locomotive  having  single  expansion  engines.  This 
may  be  expressed  in  terms  of  the  dimensions  of  the  cylin- 
ders and  the  unit  steam  pressure.  Let  p  be  the  unit 
steam  pressure  in  pounds  per  square  inch,  I  the  length  of 
the  cylinder  in  inches,  d  the  diameter  of  the  cylinder  in 
inches,  and  d1  the  diameters  of  the  drivers  in  inches  ;  then 


dl 

For  uniform  motion  the  train  resistance  cannot  exceed 
the  friction  force  or  force  of  adhesion  between  the  drivers 
and  the  rails,  since  these  are  the  external  horizontal  forces 
acting  on  the  engine  at  any  time.  This  force  of  adhesion 
in  American  practice  is  usually  taken  as  ^  or  ^  of  the 
weight  on  the  drivers. 

Problem  367.  Derive  equation  (3)  of  this  article  by  considering 
the  work  done  on  the  whole  engine. 


282  APPLIED  MECHANICS  FOB  ENGINEERS 

Problem  368.  What  resistance  R  may  be  overcome  by  a  locomo- 
tive moving  at  uniform  speed,  diameter  of  drivers  62  in.,  cylinders 
16  x  24  in.,  and  a  steam  pressure  on  the  piston  of  160  Ib.  per  square 
inch  ?  What  should  be  the  weight  of  the  locomotive  on  the  drivers? 

Problem  369.  If  the  diameter  of  the  drivers  of  a  locomotive  is 
68  in.,  and  the  size  of  the  cylinder  is  20  x  24  in.,  what  train  resistance 
may  be  overcome  by  a  steam  pressure  of  160  Ib.  per  square  inch? 

Problem  370.  A  locomotive  has  a  weight  of  155  tons  on  the  drivers. 
If  the  adhesion  is  taken  as  £,  this  allows  31  tons  for  the  drawbar  pull. 
The  train  resistance  per  ton  of  2000  Ib.,  for  a  speed  of  60  mi.  per  hour, 
is  20  Ib.  Find  the  weight  of  the  train  that  can  be  pulled  by  the 
locomotive  at  the  speed  of  60  mi.  per  hour. 

Problem  371.  An  80-car  freight  train  is  to  be  pulled  by  a  single 
expansion  locomotive  at  the  rate  of  30  mi.  per  hour.  The  weight  of 
each  car  is  60,000  Ib.,  and  the  resistance  for  this  speed  is  10  Ib.  per 
ton.  What  must  be  the  weight  on  the  drivers,  if  the  adhesion  is  £? 


CHAPTER   XIII 
FRICTION 

154.  Friction.  —  When  one  body  is  made  to. slide  over 
another,  there  is  considerable  resistance  offered  because  of 
the  roughness  of  the  two  bodies.     A  book  drawn  across 
the  top  of  a  table  is  resisted  by  the  roughness  of  the  two 
bodies.     The  rough  parts  of  the  book  sink  into  the  rough 
parts  of  the  table  so  that  when  one  of  the  bodies  tends  to 
move  over  the  other,  the  projections  interfere  and  tend  to 
stop  the   motion.     The   bearings  of   machines  are  made 
very  smooth,  and  usually  we  do  not  think  of  such  surfaces 
as  having  projections.     Nevertheless  they  are  not  perfectly 
smooth,  and  when  one  surface  is  rubbed  over  the  other,  re- 
sistance must  be  overcome.     This  resisting  force  to  the 
motion  of   one  body  over   another  is  known  as  friction. 
When  the  bodies  are  at  rest  relative  to  each  other,  the 
friction  is  known  as  the  friction  of  rest,  or  static  friction. 
When  they  are  in  motion  with  respect  to  each  other,  the 
friction  is  known  as  i\\Q  friction  of  motion,  or  kinetic  friction. 

155.  Coefficient  of  Friction.  —  If  the  body  represented  in 
Fig.  224  be  pulled  along  the  horizontal  plane  by  the  force 
P,  the  following  forces  will  be  acting  on  it  :  the.downward 
force  Gf-  and  the  reaction  R  inclined  back  of  the  vertical 
through  the  angle  0.     The  reaction  R  of  the  plane  on  the 
body  may  be  resolved  into  two  components,  one  horizontal 


284 


APPLIED  MECHANICS  FOR   ENGINEERS 


and  one  vertical.     The  horizontal  force  is  known  as  the 
force  of  friction,  and  the  normal  force,  the  normal  pressure. 

The  tangent  of  the  angle 


N 


224 


0,    or  — ,    is    called    the 

>-P      coefficient  of  friction.   This 

coefficient      of      friction, 

which  we  shall  represent 
by  /,  may  be  defined  as 
the  ratio   of  the  force  of 
friction  to  the  normal  pressure  ;  it  is  an  absolute  number. 

The  angle  6,  between  R  and  the  normal  to  the  surface 
of  contact,  is  called  the  angle  of  friction. 

(9=  tan"1/- 

The  coefficient  of  friction  is  usually  determined  by  al- 
lowing a  body  to  slide  down  an  inclined  plane,  as  shown  in 
Fig.  225.  The  angle  0  is  in- 
creased until  the  force  of  friction 
F  will  just  keep  the  body  from 
sliding  down  the  plane.  The 
angle  6  is  then  called  the  angle  of 
repose,  and  the  tangent  of  6  is  the 
coefficient  of  friction. 

Proof:  When  the  body  is  just 
on  the  point  of  slipping  down, 
the  force  R  must  just  balance  Cr.  Hence  the  angle  be- 
tween R  and  N  is  equal  to  the  angle  of  inclination  of 
the  plane,  or  the  angle  of  friction  is  equal  to  the  angle  of 
inclination  of  the  plane. 


N 


FRICTION  285 

It  is  possible  with  such  an  apparatus  to  determine  the 
coefficient  of  friction  for  various  materials.  It  has  been 
found  that  after  motion  begins  the  friction  is  less;  that  is, 
the  friction  of  motion  is  less  than  the  friction  of  rest.  This 
is  an  important  law  for  engineers. 

156.  Laws  of  Friction  for  Dry  Surfaces.  —  Very  little  was 
known  of  the  laws  of  friction  until  within  the  last  seventy- 
five  years.  About  1820  experiments  were  made  that 
seemed  to  show  that,  for  such  materials  as  wood,  metals,  etc., 
friction  varies  with  the  pressure,  and  is  independent  of 
the  extent  of  the  rubbing  surfaces,  the  time  of  contact, 
and  the  velocity.  A  little  later  (1831)  Morin  published 
the  following  three  laws  as  a  result  of  his  experiments  on 
friction : 

(1)  The  friction   between  two  bodies  is  directly  propor- 
tional to  the  pressure;  that  is,  the  coefficient  of  friction  is 
constant  for  all  pressures. 

(2)  The  coefficient  and  amount  of  friction  for  any  given 
pressure  is  independent  of  the  area  of  contact. 

(3)  The  coefficient  of  friction  is  independent  of  the   ve- 
locity, although  static  friction  is  greater  than  kinetic  fric- 
tion. 

These  laws  of  Morin  hold  approximately  for  dry,  unlu- 
bricated  surfaces,  although  it  has  been  found  that  an  in- 
crease in  speed  lowers  the  coefficient  of  friction.  The 
coefficient  of  friction  is  a  little  greater  for  light  pres- 
sures upon  large  areas  than  for  great  pressures  on  small 
areas. 

The  following  is  a  table  of  some  of  the  coefficients  of 
friction  as  determined  by  Morin  : 


286  APPLIED  MECHANICS  FOR  ENGINEERS 


COEFFICIENTS  OF  FRICTION,  DUE  TO  MORIN 


MATERIAL 

CONDITION  OF  SURFACE 

COEFFICIENT 

OF 

FRICTION 

ANGLE 

OF 

FRICTION 

Brick  on  limestone 

Dry 

.67 

33°  50' 

Cast  iron  on  cast  iron 

Slightly  greased 

.16 

9°    6' 

Cast  iron  on  oak 

Wet 

.65 

33°   2' 

Copper  on  oak 

.17 

9°  38' 

Copper  on  oak 

Greased 

.11 

6°  17' 

Leather  on  cast  iron 

.28 

15°  39' 

Leather  on  cast  iron 

Wet 

.38 

20°  49' 

Leather  on  cast  iron 

Oiled 

.12 

6°  51' 

Leather  on  oak 

Fibers  parallel 

.74 

36°  30' 

Leather  on  oak 

Fibers  crossed 

.47 

25°  11' 

Oak  on  oak 

Fibers  parallel,  dry 

.62 

31°  48' 

Oak  on  oak 

Fibers  crossed,  dry 

.54 

28°  22' 

Oak  on  oak 

Fibers  parallel,  soaped 

.44 

23°  45' 

Oak  on  oak 

Fibers  crossed,  wet 

.71 

35°  23' 

Oak  on  oak 

Fibers  end  to  side,  dry 

.43 

23°  16' 

Oak  on  oak 

Fibers  parallel,  greased 

.07 

4°    6' 

Oak  on  oak 

Heavily  loaded,  greased 

.15 

8°  45' 

Oak  on  pine 

Fibers  parallel 

.67 

33°  50' 

Oak  on  limestone 

Fibers  on  end 

.63 

32°  15' 

Oak  on  hemp  cord 

Fibers  parallel 

.80 

38°  40' 

Pine  on  pine 

Fibers  parallel 

.56 

29°  15' 

Pine  on  oak 

Fibers  parallel 

.53 

27°  56' 

Wrought  iron  on  oak 

Wet 

.62 

31°  48' 

Wrought  iron  on  oak 

.65 

33°   2' 

Wrought  iron  on  wrought  iron 

.28 

15°  39' 

Wrought  iron  on  cast  iron 

.19 

10°  46' 

Wrought  iron  on  limestone 

.49 

26°   7' 

Wood  on  metal 

Greased 

.10 

6°   0' 

Wood  on  smooth  stone 

Dry 

.58 

30°   7' 

Wood  on  smooth  earth 

Dry 

.33 

18°  16' 

FRICTION 


287 


Problem  372.  Find  the  force  P  necessary  to  move  with  uniform 
velocity  a  weight  of  100  Ib.  up  a  plane  inclined  30°  to  the  horizontal 
(a)  when  P  is  horizontal,  (b)  when  parallel  to  the  plane,  (c)  when  in- 
clined at  an  angle  of  60°  to  the  horizontal,  given  that  the  coefficient  of 
friction  between  the  weight  and  the  plane  is  .20.  Find  the  force  just 
necessary  to  prevent  the  body  from  sliding  down  the  plane  in  each  of 
the  cases. 


FIG.  226 

Problem  373.  Show  that  the  force  P,  inclined  at  an  angle  <f>  to 
the  plane,  that  will  (a)  just  move  the  weight  up  the  plane,  (6)  just 
prevent  it  from  sliding  down  the  plane  (Fig.  226),  is 


n  - 
" 


sin 


where  0  is  the  angle  of  friction. 

Problem  374.     Show  that  the  least  values  of  P  in  the  preceding 
problem  are  when  <f>  —  0  in  (a),  and  when  <f>  =  —  0  in  (6)  ;  i.e.  when 

(a)  P  =  G  sin  (a  +  0),         (b)  P  =  G  sin(a  -  6). 


Problem  375.  In  Fig.  227  the  weight 
G  is  raised  by  the  horizontal  force  P.  If 
the  only  friction  is  between  the  surfaces 
of  the  wedge  and  the  weight,  prove  that 
the  value  of  P  just  sufficient  to  raise  the 
weight  is 


LJ 


FIG.  227 


288 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  376.     Defining  the  efficiency,  E,  of  the  wedge  as  the  ratio 
of  the  useful  work  accomplished  in  raising  the  weight  to  the  total 
work  done  by  P  (Fig.  227)  show  that 
-p  _       tan  a 
~  tan  (a  +  0) 

Problem  377.     For  a  given  value  of  0,  show  that  E  is  a  maximum 
when 


(Since  a  square-threaded  screw  may  be  regarded  as  an  inclined  plane, 

this  formula  also  holds  for  such  a 
screw.) 


Problem  378.  Find  the  value 
of  the  horizontal  force  P  that  will 
just  raise  the  weight  of  500  Ib.  in 
Fig.  228,  given  that  the  coefficients 
of  friction  at  A,  B,  and  C  are  re- 
spectively .20,  .25,  and  .30. 


SUGGESTION.  Consider  the  wedge 
and  block  separately,  and  the  forces  that  hold  them  in  equilib- 
rium. 

157.  Friction  of  Lubricated  Surfaces.  —  The  laws  of  fric- 
tion, as  given  by  Morin  and  stated  in  the  preceding  article, 
hold  approximately  for  rubbing  surfaces,  when  the  sur- 
faces are  dry  or  nearly  so  ;  that  is,  for  poorly  lubricated 
surfaces.  If,  however,  the  surfaces  are  well  lubricated  so 
that  the  projections  of  one  do  not  fit  into  the  other,  but 
are  kept  apart  by  a  film  or  layer  of  the  lubricant,  the  laws 
of  Morin  are  not  even  approximately  true.  The  study  of 
the  friction  of  lubricated  surfaces,  then,  may  be  divided 
into  two  parts  :  (1)  the  study  of  poorly  lubricated  bear- 
ings, and  (2)  the  study  of  well  lubricated  bearings,  the 


FRICTION  289 

friction  of  which  varies  from  ^  to  y1^  that  of  dry  or  poorly 
lubricated  bearings. 

Since  the  friction  of  poorly  lubricated  bearings  is  about 
the  same  as  that  of  dry  surfaces,  we  shall  consider  that 
the  laws  of  Morin  hold,  and  shall  confine  our  attention  to 
the  friction  of  well  lubricated  bearings.  If  the  lubricant 
is  an  oil,  the  friction  of  the  bearing  is  no  longer  due  to 
one  surface  rubbing  over  the  other,  but  to  the  friction 
between  the  bearing  and  the  oil,  and  to  the  internal  fric- 
tion of  the  oil.  That  is,  the  oil  adheres  to  the  two  sur- 
faces, and  its  own  particles  attract  each  other,  and  the 
motion  of  one  of  the  surfaces  with  respect  to  the  other 
changes  the  positions  of  the  oil  particles.  It  is  to  be 
expected,  then,  that  the  friction  of  an  oiled  bearing  will 
depend  upon  the  viscosity  of  the  oil,  upon  the  thickness  of 
the  layer  interposed  between  the  surfaces,  and  upon  the 
velocity  and  form  of  the  bearing. 

The  coefficient  of  friction  is  no  longer  constant,  but 
varies  with  the  temperature,  velocity,  and  pressure.  The 
variation  of  the  coefficient  of  friction  of  a  paraffine  oil 
with  temperature  is  shown  in  Fig.  229  when  the  pressure  on 
the  bearing  is  33  Ib.  per  square  inch  and  a  velocity  of  rub- 
bing of  296  ft.  per  minute.  It  is  seen  that  the  coefficient 
of  friction  decreases  with  increase  of  temperature  until  a 
temperature  of  80°  F.  is  reached,  when  it  increases  rapidly. 
This  means  that  above  this  temperature  the  oil  is  so  thin 
that  it  is  squeezed  out  of  the  bearing,  and  the  conditions 
of  dry  bearing  are  approached.  The  temperature  at 
which  oils  show  an  increasing  coefficient  of  friction  is  dif- 
ferent for  different  oils,  even  at  the  same  pressure  and 


290 


APPLIED  MECHANICS  FOR  ENGINEERS 


velocity.  The  curve  in  Fig.  229,  however,  may  be  re- 
garded as  typical  of  all  oils  when  the  pressure  and  velocity 
are  constant. 


COEFFICIENT  OF  FRICTION 
OF  A  PARAFFINE  OIL 

PRESSURE  83  LBS.  PER  SQ.  INCH 
VELOCITY  296  FEET  PER  MINUTE 


.02  .03 

COEFFICIENT  OF  FRICTION 

FIG.  229 


The  following  table,  due  to  Professor  Thurston,  shows 
the  relation  between  the  coefficient  of  friction  and  tem- 
perature for  a  sperm  oil  in  steel  bearings  when  the  veloc- 
ity of  rubbing  is  30  ft.  per  minute: 


FRICTION 


291 


PRESSURE,  LB. 
PER  SQ.  IN. 

TEMPERATURE, 
DEGREES  F. 

COEFFICIENT 
OF  FRICTION 

PRESSURE,  LB. 
PER  SQ.  IN. 

TEMPERATURE, 
DEGREES  F. 

COEFFICIENT 
OF  FRICTION 

200 

150 

.0500 

100 

110 

.0025 

200 

140 

.0250 

50 

110 

.0035 

200 

130 

.0160 

4 

110 

.0500 

200 

120 

.0110 

200 

90 

.0040 

200 

110 

.0100 

150 

90 

.0025 

200 

100 

.0075 

100 

90 

.0025 

200 

95 

.0060 

50 

90 

.0035 

200 

90 

.0056 

4 

90 

.0400 

150 

110 

.0035 

It  is  seen  that  for  a  pressure  of  200  Ib.  per  square  inch 
as  the  temperature  increases  from  90°  F.  the  coefficient  in- 
creases, indicating  that  the  temperature  of  90°,  for  the 
given  pressure  and  velocity,  was  above  the  temperature 
at  which  the  oil  became  so  thin  as  to  be  squeezed  out  and 
the  bearing  to  approach  the  condition  of  a  dry  bearing. 
For  a  constant  temperature  110°  F.  and  90°  F.  the  coeffi- 
cient is  seen  to  decrease  with  increase  of  pressure  up  to  a 
certain  point  and  then  to  increase.  This  is  a  typical  be- 
havior of  oils  when  the  temperature  is  constant  and  the 
pressure  varies. 

At  speeds  exceeding  100  ft.  per  minute,  the  same  author- 
ity found  "  that  the  heating  of  the  bearings  within  the 
above  range  of  temperatures  decreases  the  resistance  due 
to  friction,  rapidly  at  first  and  then  slowly,  and  gradually 
a  temperature  is  reached  at  which  increase  takes  place  and 
progresses  at  a  rapidly  accelerating  rate." 

The  relation  between  the  coefficients  of  rest  and  of  motion 
as  determined  by  Professor  Thurston  for  three  oils  is  given 


292 


APPLIED  MECHANICS  FOR   ENGINEERS 


below.    The  journals  were  cast  iron,  in  steel  boxes  ;  velocity 
of  rubbing  150  ft.  per  minute  and  a  temperature  115°  F. 


SPERM  OIL 

WEST  VIRGINIA  OIL 

LARD 

PRESSURE, 
LB.  PER 
SQ.  IN. 

At  150 
ft.  per 
min. 

At 
start- 
ing 

At 
stop- 
ping 

At  150 
ft.  per 
min. 

At 
start- 
ing 

At 
stop- 
ping 

At  150 
ft.  per 
min. 

At 
start- 
ing 

At 
stop- 
ping 

50 

.013 

.07 

.03 

.0213 

.11 

.025 

.02 

.07 

.01 

100 

.008 

.135 

.025 

.015 

.135 

.025 

.0137 

.11 

.0225 

250 

.005 

.14 

.04 

.009 

.14 

.026 

.0085 

.11 

.016 

500 

.004 

.15 

.03 

.00515 

.15 

.018 

.00525 

.10 

.016 

750 

.0043 

.185 

.03 

.005 

.185 

.0147 

.0066 

.12 

.020 

1000 

.009 

.18 

.03 

.010 

.18 

.017 

.0125 

.12 

.019 

Steel  Journals  and  Brass  Boxes 


500 
1000 

.0025 
.008 

.004 
.009 

It  is  seen  that  the  coefficient  of  friction  at  starting  is  much 
greater  than  at  stopping,  and  that  these  are  both  much 
greater  than  the  value  at  a  speed  of  150  ft.  per  minute. 

For  an  intermittent  feed  such  as  is  given  by  one  oil  hole, 
without  a  cup,  oiled  occasionally,  Professor  Thurston  found 
for  steel  shaft  in  bronze  bearings,  with  a  speed  of  rubbing 
of  720  ft.  per  minute,  the  following  coefficipnts  of  friction  : 


OIL 


PRESSURE,  LB.  PER  SQ.  IN. 


8 

16 

33 

48 

Sperm  and  lard  .... 
Olive  and  cotton  seed 
Mineral  oils 

.159-.25 
.160-.283 
154-  261 

.138-.192 
.107-.245 
.145-.233 

.086-.141 
.101-.168 
.086-.178 

.077-.144 
.079-.131 
.094-.222 

FRICTION  293 

The  results  show  that  the  coefficient  decreases  with  the 
pressure  within  the  range  reported,  but  that  the  results 
are  considerably  higher  than  those  for  well  lubricated 
bearings.  He  also  found  in  connection  with  the  same 
tests  that  with  continuous  lubrication  sperm  oil  gave  the 
following  coefficients : 

PRESSURE,  COEFFICIENT 

LB.  PEE  SQ.  IN.  OF  FRICTION 

50  .0034 

200  .0051 

300  .0057 

The  results  of  tests  of  the  friction  of  well-lubricated 
bearings  are  summarized  by  Goodman  (Engineering  News, 
April  7  and  14,  1888)  as  follows  : 

(a)  The  coefficient  of  friction  of  well  lubricated  surfaces  is 
from  ^  to  y1-^  that  of  dry  or  poorly  lubricated  surfaces. 

(£>)  The  coefficient  of  friction  for  moderate  pressures  and 
speeds  varies  approximately  inversely  as  the  normal  pressure ; 
the  frictional  resistance  varies  as  the  area  in  contact,  the  nor- 
mal pressure  remaining  the  same. 

(c)  For  low  speeds  the  coefficient  of  friction  is  abnormally 
high,  but  as  the  speed  of  rubbing  increases  from  about  10  to 
100  ft.  per  minute,  the  coefficient  of  friction  diminishes,  and 
again  rises  when  that  speed  is  exceeded,  varying  approxi- 
mately as  the  square  root  of  the  speed. 

(c?)  The  coefficient  of  friction  varies  approximately  in- 
versely as  the  temperature,  within  certain  limits;  namely, 
just  before  abrasion  takes  place. 

158.  Method  of  Testing  Lubricants. —  To  make  the  matter 
of  the  tests  of  the  friction  of  lubricants  clear,  it  will  be 


294 


APPLIED  MECHANICS   FOR  ENGINEERS 


convenient  to  make  use  of  the  description  of  a  testing 
machine  used  by  Dean  W.  F.  M.  Goss  at  Purdue  Uni- 
versity on  graphite,  and  a  mixture  of  graphite  and  sperm 


FIG.  230 

oil.  In  making  the  tests  the  apparatus  shown  in  Figs.  230 
and  231  was  used.  (See  "  A  Study  in  Graphite,"  Joseph 
Dixon  Crucible  Co.) 

This  apparatus  represents,  in  principle,  the  machines 
generally  used  for  testing  lubricants.  It  is  therefore 
shown  in  some  detail.  The  weight  Gr  is  hung  from  the 
shaft  upon  which  it  is  suspended  by  the  form  of  box  to  be 
tested.  The  •  desired  speed  of  rubbing  is  obtained  by 
means  of  the  cone  of  pulleys,  and  the  pressure  on  the  bear- 
ing is  adjusted  by  the  spring.  The  temperature  of  the 
bearing  is  read  from  the  thermometer  inserted  in  the  bear- 
ing. When  rotation  takes  place,  the  weight  Gr  is  rotated 
a  certain  distance  dependent  upon  the  friction.  This  dis- 
tance is  measured  on  the  scale  A.  The  forces  acting  upon 


FRICTION 


295 


the  pendulum  Gr  are  shown  in  Fig.  231,  where  R  represents 
the  resistance  of  the  spring,  F  the  force  of  friction,  I  the 
distance  of  the  center  of  gravity  of  Gr 
from  the  axis  of  rotation,  <f>  the  angle 
through  which  Gr  is  deflected,  r  the 
radius  of  the  shaft,  and  /  the  coeffi- 
cient of  friction.  Taking  moments 
about  the  center  of  the  shaft,  we  have, 
when  Gr  is  held  in  the  position  shown, 
due  to  the  friction, 


or 


if(JR 


=-  sn 


G) 


FIG.  231 


It   is   customary   to   take    Gr   small 
compared  with  R,  so  that  the  pressure 
on  both  sides  of  the  bearing  may  be  considered  equal  to 
jR,  the  resistance  of  the  spring.     The  formula  then  becomes 


/  = 


2rjR 


The  spring  is  easily  calibrated  so  that  R  may  be  made  any- 
thing desired  by  compressing  the  spring  through  the  ap- 
propriate distance,  as  indicated  on  the  scale  V  (Fig.  230). 
The  quantities  6r,  ?,  r,  and  R  are  known,  and  <f>  can  be 
read  so  that /can  be  calculated. 

The  results  of  tests  made  upon  a  mixture  of  graphite 
and  oil  as  a  lubricant  are  given  in  the  pamphlet.  The 
tests  were  run  under  200  Ib.  per  square  inch  pressure,  at 
a  speed  of  rubbing  of  145  ft.  per  minute.  Oil  was  dropped 


296 


APPLIED  MECHANICS  FOR   ENGINEERS 


into  the  bearing  at  the  rate  of  about  12  drops  per  minute, 
showing  a  coefficient  of  friction  of  \. 

Problem  379.  If  the  weight  of  the  pendulum  is  360  lb.,  the 
diameter  of  the  shaft  4£  in.,  distance  of  the  center  of  gravity  of  G 
from  the  center  of  shaft  2  ft.,  the  angle  <£  5  degrees,  and  the  average 
resistance  of  the  spring  1000  lb.,  find  the  coefficient  of  friction.  The 
weight  G  should  not  be  neglected  in  this  case. 

159.  Rolling  Friction.  —  The  resistance  offered  to  the 
rolling  of  one  body  over  another  is  known  as  rolling  fric- 
tion. It  is  entirely  different  from  sliding  friction,  and  its 

laws  are  not  so  well 
understood.  When  a 
wheel  or  cylinder  (Fig. 
232)  rolls  over  a  track, 
the  track  is  depressed 
and  the  wheel  dis- 
torted. The  force  P 
necessary  to  overcome  this  depression  and  distortion  is 
equal  to  the  rolling  friction. 

The  forces  acting  on  the  wheel  are  seen  from  Fig.  232 
to  be:  P  the  working  force,  IP" the  weight  on  the  wheel, 
and  R  the  reaction  of  the  track  or  roadway.  This  upward 
pressure  R  is  not  quite  vertical,  but  has  its  point  of  ap- 
plication a  short  distance  K'  from  the  vertical.  Its  line 
of  action  passes  through  the  center  of  the  wheel.  The 
distance  K'  depends  chiefly  upon  the  roadway  ;  it  is  called 
the  coefficient  of  rolling  friction.  It  is  measured  in  inches 
and  is  not  a  coefficient  of  friction  in  the  strict  sense  that/ 
is  the  coefficient  of  sliding  friction. 

Taking  moments  about  the  point  of  application  of  R^ 


FIG.  232 


FRICTION  297 

we  have,  approximately, 

WK'  =  Pr, 

so  that  x/  =       ,orl>  =          . 


When  the  track  or  roadway  is  elastic  or  nearly  so,  we 
have  a  condition  something  like  that  represented  in  Fig. 
233.      The    wheel 
sinks  into  the  ma- 
terial and  pushes  it 
ahead,  at  the  same 

time  it   comes    up  f\  -  J          >  p 

behind  the  wheel. 
For  a  portion  of 
the  wheel  on  each 
side  of  the  point  0 
the  roadway  is  simply  compressed  ;  over  the  remainder  of 
the  surface  in  contact,  however,  slipping  occurs,  as  indi- 
cated by  the  arrows.  The  resultant  resistance,  however, 
is  in  front  of  the  vertical  through  the  center,  and  we  have, 
as  in  the  case  of  imperfectly  elastic  roadways, 


It  has  been  found  by  Reynolds  (See  Phil.  Trans.  Royal 
Soc.,  Vol.  166,  Part  1)  that  when  a  cast-iron  roller  rolls 
on  a  rubber  track,  the  slippage,  due  to  the  elasticity  of  the 
track,  may  amount  to  as  much  as  .84  in.  in  34  in.  An 
elastic  roller  rolling  on  a  hard  track  will  roll  less  than  the 
geometrical  distance  traveled  by  a  point  on  the  circumfer- 
ence. When  the  roller  and  track  are  of  the  same  material, 
the  roller  rolls  through  less  than  its  geometrical  distance. 


298  APPLIED  MECHANICS  FOR  ENGINEERS 

160.    Antifriction  Wheels.  —  The  axle  A,   of  radius   r, 
carrying  a  weight  W,  rests  upon  two  wheels  of  radius  r2, 

turning  on  axles  of  radius  ^  (Fig. 
234). 

The  force  on  each  of  the  bear- 
ings of  wheels  B  and  C  is 

W 


2  cos  ft' 

and  if  F  is  the  friction  at  each  of 
the  bearings  of  B  and   (7,  and  / 
FIG.  234  the  coefficient  of  sliding  friction, 

W 


and  the  work  lost  in  friction  at  the  bearings  of  B  and  0  in 
one  revolution  of  the  axle  A  is 


cos/3    L    r2 

Since  the  axle  A  rolls  on  B  and  (7,  there  is  no  sliding  fric- 
tion there  and  the  rolling  friction  is  small  enough  to  be 
neglected. 

If  A  were  in  an  ordinary  bearing,  the  work  lost  per 
revolution  would  be 

Therefore  the  ratio  of  work  lost  with  antifriction  wheels 
to  the  work  lost  with  plain  bearing  is 

r2  cos  ft 

This  ratio  decreases  as  the  ratio  r^Jr^  increases  and  as  ft 
decreases. 


FRICTION  299 

Problem  380.  If  W  =  4  tons,  the  radius  of  the  shaft  is  2  in., 
and  the  coefficient  of  friction  is  .07,  what  work  is  lost  per  revolution? 
If  the  shaft  makes  3  revolutions  per  second,  what  horse  power  is  lost 
in  friction  ?  Given  also  /?  =  45°,  rl  =  f  in.,  and  r2  =  4  in. 

Problem  381.  In  the  case  of  the  shaft  mentioned  in  the  preced- 
ing problem,  how  much  more  horse  power  would  it  take  if  the  bear- 
ing were  plain  ?  What  value  of  ft  would  give  the  same  loss  due  to 
friction  in  both  the  plain  bearing  and  the  one  provided  with  friction 
wheels  ? 

161.  Resistance  of  Ordinary  Roads.  —  Resistance  to  trac- 
tion consists  of  axle  friction,  rolling  friction,  and  grade 
resistance.  Axle  friction  varies  from  .012  to  .02  of  the 
load,  for  good  lubrication,  according  to  Baker.  The 
tractive  power  necessary  to  overcome  axle  friction  for 
ordinary  American  carriages  has  been  found  to  be  from 
3  Ib.  to  3^  Ib.  per  ton,  and  for  wagons  with  medium-sized 
wheels  and  axles  from  3|  Ib.  to  4J  Ib.  per  ton. 

The  total  tractive  force  per  ton  of  load,  for  wheels  50  in., 
30  in.,  and  26  in.,  in  diameter,  respectively,  is,  according 
to  Baker  (Engineering  News,  March  6,  1902) : 


TRACTIVE  FORCE 

IN  POUNDS 

On  macadam  roads 

57 

61 

70 

On  timothy  and  blue 

grass  sod,  dry,  grass  cut  .     . 

132 

145 

179 

On  timothy  and  blue 

grass  sod,  wet  and  springy   . 

173 

203 

288 

On  plowed  ground,  not  harrowed,  dry  and  cloddy  . 

252 

303 

374 

Rolling  resistance  is  influenced  by  the  width  of  the  tire. 
According  to  Baker,  poor  macadam,  poor  gravel,  compres- 
sible earth  roads,  and,  on  agricultural  lands,  narrow  tires, 


300 


APPLIED  MECHANICS  FOR  ENGINEERS 


usually  require  less  traction.  On  earth  roads  composed  of 
dry  loam  with  2  to  3  in.  of  loose  dust,  traction  with  IJ-in. 
tires  was  90  Ib.  per  ton,  and  with  6-in.  tires  106  Ib.  per 
ton.  On  the  same  road  when  it  was  hard  and  dry,  with  no 
dust,  that  is,  when  it  was  compressible,  the  traction  was 
found  to  be  149  Ib.  per  ton  with  IJ-in.  tires  and  109  Ib. 
per  ton  with  6-in.  tires.  On  broken  stone  roads,  hard  and 
smooth,  with  no  dust  or  loose  stones,  the  traction  per  ton 
was  121  Ib.  with  IJ-in.  tires,  and  98  Ib.  with  6-in.  tires. 
Moisture  on  the  surface  or  mud  increases  the  traction. 

Morin  found  that  with  44-in.  front  and  54-in.  rear 
wheels  on  hard  dry  roads  the  traction  per  ton  was  114  Ib. 
with  either  l|-in.  or  3-in.  tires.  On  wood-block  pave- 
ments the  traction  per  ton  was  28  Ib.  with  IJ-in.  tires, 
and  38  Ib.  with  6-in.  tires. 

On  asphalt,  bricks,  granite,  macadam,  and  steel-road 
surfaces,  investigated  by  Baker,  the  traction  per  ton 
varied  from  17  Ib.  to  70  Ib.,  the  average  being  38  Ib. 

Morin  gives  the  coefficient  of  rolling  friction  for  wagons 
on  soft  soil  as  .065  in.,  and  on  hard  roads  .02  in.  Accord- 
ing to  Kent  ("  Pocket-Book  "),  tests  made  upon  a  loaded 
omnibus  gave  the  following  results : 


PAVEMENT 

SPEED.  MILES 
PER  HOUR 

COEFFICIENT, 
INCHES 

RESISTANCE,  PER 
TON,  IN  LB. 

Granite 

2  87 

007 

1741 

Asphalt 

3  56 

0121 

2714 

Wood     

3.34 

.0185 

41.60 

Macadam,  graveled      .     .     . 
Macadam,  granite,  new    .     . 

3.45 
3.51 

.0199 
.0451 

44.48 
101.09 

FRICTION 


301 


Problem  382.  Compare  the  resistance  offered  to  a  load  of  two 
tons  pulled  over  asphalt,  macadam,  good  earth  roads,  or  wood-block 
pavement.  Width  of  tires,  6  in. 

162.  Roller  Bearings.  —  In  the  roller  bearings  the  shaft 
rolls  on  hardened  steel  rollers  as  shown  in  cross  section  in 
Fig.  235.  The  roll- 
ers are  kept  in  place 
in  some  way  similar 
to  that  shown  in  the 
journal  of  Fig.  236. 
Such 
used 


FIG.  235 


bearings    are 

where  heavy 
loads  are  to  be  car- 
ried. Tests  of  roller 
bearings  have  been  made  by  Dean  C.  H.  Benjamin 
(Machinery,  October,  1905),  who  determined  the  follow- 
ing values  for  the  coefficient  of  friction.  Speed  480 
revolutions  per  minute. 


DIAMETER  OF 
JOUENAL,  IN  INCHES 

ROLLER  BKARINU 

PLAIN  CAST-IRON  BEARING 

Max. 

Min. 

Average 

Max. 

Min. 

A.verage 

IH 

.036 

.019 

.026 

.160 

.099 

.117 

2t 

.052 
.041 

.034 
.025 

.040 
.030 

.129 
.143 

.071 
.076 

.094 
.104 

2H 

.053 

.049 

.051 

.138 

.091 

.104 

It  was  found  that  the  coefficient  of  friction  of  roller 
bearings  is  from  ^  to  ^  that  of  plain  bearings  at  moderate 
speeds  and  loads.  As  the  load  on  the  bearing  increased, 


302 


APPLIED  MECHANICS  FOR  ENGINEERS 


the  coefficient  of  friction  decreased.     Tightening  the  bear- 
ing was  found  to  increase  the  friction  considerably. 

Tests  of  the  friction  of  steel 
rollers  1,  2,  3,  and  4  in.  in  diameter 
are  reported  in  the  Trans.  Am. 
Soc.  C.  E.,  August,  1894.  The 
rollers  were  tested  between  plates 
1|  in.  thick  and  5  in.  wide,  ar- 
ranged as  shown  in  Fig.  237. 
Tests  were  made  with  the  plates 
and  rollers  of  cast  iron,  wrought 
iron,  and  steel.  The  friction  P1  for 

.0063 
Vr 


FIG.  236 


unit  load  P  was  found  to  be 


.0120 


for  cast-iron  rollers  and  plates,—    —for  wrought  iron,  and 

Vr 

f)Q7Q 

'- — —  for  steel,  where  r  represents  the  radius  of  the  roller 

Vr 

in  inches.  The  rollers  were 
turned  and  the  plates  planed, 
but  neither  was  polished. 

163.     Ball     Bearings.  —  For 

high  speeds  and  light  or  moder- 
ate loads  the  friction  is  much 
reduced  by  the  use  of  hardened 
steel  balls  instead  of  the  steel 
rollers.  These  bearings  are  now 

used  on  all  classes  of  machinery,  giving  a  much  greater 
efficiency  except  for  heavy  loads.  The  principal  objection 
to  the  ball  bearing  seems  to  be  due  to  the  fact  that  there  is 


FRICTION 


303 


FIG.  238 


so  little  area  of  contact  between  the  balls  and  bearing 
plates.  This  gives  rise  to  very  high  stresses  over  these  areas, 
and  consequently  a  considerable 
deformation  of  the  balls.  When 
the  ball  has  been  changed  from  its 
spherical  form,  it  is  no  longer  free 
to  roll,  and  the  friction  increases 
rapidly.  Some  authorities  con- 
sider a  load  of  from  50  to  150  Ib. 
sufficient  for  balls  varying  in  size 
from  J  to  J  inch  in  diameter.  Fig- 
ure 238  illustrates  a  type  of  bear- 
ing used  for  shafts,  and  Fig.  239  a  type  used  for  thrust  blocks. 
The  conclusions  reached  by  Goodman  from  a  series  of 
tests  on  bicycle  bearings  (Proc.  Inst.  C.  E.,  Vol.  89)  are 

as  follows : 

(1)  The  coefficient  of  friction 
of  ball  bearings  is    constant  for 
varying  loads,  hence  the  frictional 
resistance  varies  directly  as  the 
load. 

(2)  The  friction  is  unaffected 
by  a  change  of  temperature. 

The  bearings  were  oiled  be- 
fore starting  the  tests.  The 
coefficient  of  friction  for  ball 
bearings  was  found  to  be  rather  higher  than  for  plain  bear- 
ings with  bath  lubrication,  but  lower  than  for  ordinary 
lubrication.  Ball  bearings  will  also  run  easily  with  a  less 
supply  of  oil.  The  following  table  gives  the  results  of 


FIG.  239 


304 


APPLIED  MECHANICS  FOR   ENGINEERS 


tests  of  ball  bearings.     The  bearings  were  oiled  before  start- 
ing, and  the  tests  were  run  at  a  temperature  of  68°  F. 


LOAD  ON 
BEARING 
IN  LB. 

19 

REVOLUTIONS  PER  MIN. 

157 

REVOLUTIONS  PER  MIN. 

350 

REVOLUTIONS  PER  MIN. 

Coeff.  friction 

Friction,  Ib. 

Coeff.  friction 

Friction,  Ib. 

Coeff.  friction 

Friction,  Ib. 

10 

.0060 

.06 

.0105 

.10 

.0105 

.10 

20 

.0045 

.09 

.0067 

.13 

.0120 

.24 

30 

.0050 

.15 

.0050 

.15 

.0110 

.33 

40 

.0052 

.21 

.0052 

.21 

.0097 

.39 

50 

.0054 

.27 

.0054 

.27 

.0090 

.45 

60 

.0050 

.30 

.0055 

.33 

.0075 

.45 

70 

.0049 

.34 

.0054 

.38 

.0068 

.47 

80 

.0048 

.38 

.0062 

.49 

.0060 

.48 

90 

.0050 

.45 

.0068 

.61 

.0060 

.54 

100 

.0058 

.58 

.0069 

.69 

.0057 

.57 

110 

.0054 

.59 

.0065 

.71 

.0060 

.66 

120 

.0055 

.66 

.0075 

.90 

.0057 

.68 

130 

.0058 

.75 

.0078 

1.01 

.0062 

.81 

140 

.0056 

.78 

.0077 

108 

.0060 

!84 

150 

.0060 

.90 

.0083 

1.24 

.0062 

.93 

160 

.0075 

1.20 

.0081 

1.29 

.0058 

.93 

170 

.0079 

1.34 

.0078 

1.33 

.0055 

.93 

180 

.0079 

1.42 

.0078 

1.40 

.0053 

.95 

190 

.0087 

1.65 

.0076 

1.44 

.0054 

1.03 

200 

.0090 

1:80 

.0081 

1.62 

.0060 

1.20 

Another  series  of  tests,  run  with  a  constant  load  on  the 
bearing  of  200  Ib.  and  a  temperature  of  86°  F.,  shows  the 
variation  of  the  coefficient  of  friction  with  the  speed.  It 
is  seen  that  as  the  speed  increased  the  coefficient  and  the 
friction  decreased.  The  preceding  table,  however,  shows, 
for  loads  below  175  Ib.,  an  increase  in  the  coefficient  with 
increase  in  speed.  In  particular,  this  table  shows  that  for 


FRICTION 


305 


loads  below  80  Ib.  the  coefficient  increased  with  increase 
of  speed ;  for  loads  between  90  and  175  Ib.  it  increased 
when  the  speed  was  150  r.p.m.  and  decreased  when  it  was 
350  r.p.m.  Beyond  175  Ib.  the  coefficient  increased. 


REVOLUTIONS  PEE  MINUTE 

COEFFICIENT  FRICTION 

FRICTION  POUNDS 

15 

.00735 

1.47 

93 

.00465 

.93 

175 

.00375 

.75 

204 

.00345 

.69 

280 

.00300 

.60 

It  seems  from  the  data  given  that  the  first  conclusion 
of  Goodman's  should  be  changed  to  read  :  the  coefficient  of 
friction  of  ball  bearings  is  constant  for  varying  loads,  up  to 
a  certain  limit,  beyond  which  it  increases  with  increase  of 
load.  This  limit  is  about  150  Ib.  in  the  tests  reported. 

Tests  on  ball  bearings  designed  for  machinery  subjected 
to  heavy  pressures  have  been  made  in  Germany.  (See  Zeit- 
schrift  des  Vereins  deutsche  Ingenieure,  1901,  p.  73.)  It 
was  found  that  at  speeds  varying  from  65  to  780  revolu- 
tions per  minute,  where  the  bearing  was  under  pressures 
varying  from  2200  Ib.  to  6600  Ib.,  the  coefficient  of  friction 
varied  little  and  averaged  .0015. 

Tests  of  ball  bearings  made  by  Stribeck  and  reported  by 
Hess  (Trans.  Am.  Soc.  M.  E.,  Vol.  28,  1907)  give  rise  to 
the  following  conclusions :  (a)  the  load  that  may  be  put 
upon  a  bearing  is  given  by  the  formula 

-p_  cd2n 
~  11.02' 


306 


APPLIED  MECHANICS  FOR  ENGINEERS 


where  P  is  the  load  in  pounds  on  a  bearing,  consisting  of 
one  row  of  balls,  c  is  a  constant  dependent  upon  the  mate- 
rial of  the  balls  and  supporting  surfaces  and  determined 
experimentally,  d  the  diameter  of  the  balls,  the  unit  being 
^  of  an  inch,  and  n  the  number  of  balls.  For  modern 
materials  c  varies  from  5  to  7.5.  (£)  The  coefficient  of 
friction  varied  from  .0011  to  .0095.  It  was  independent 
of  speed,  "within  wide  limits,"  and  approximated  .0015; 
this  was  increased  to  .003  when  the  load  was  about  one 
tenth  the  maximum. 

The  following  values  for  the  coefficient  of  friction  for 
heavy  loads  are  reported,  from  observation,  with  the  state- 
ment that  the  real  values  are  probably  somewhat  less : 


Revolutions 
per  minute 

65 

100 

190 

380 

580 

780 

1150 

Coefficient    of 

friction    for 

load  840  Ib. 

.0095 

.0095 

.0093 

.0088 

.0085 

.0074 

Coefficient    of 

friction   for 

load  2400  Ib. 

.0065 

.0062 

.0058 

.0053 

.0050 

.0049 

.0047 

Coefficient    of 

friction    for 

load  4000  to 

9250  Ib. 

.0055 

.0054 

.0050 

.0050 

.0041 

.0041 

.0040 

It  should  be  remembered  that  the  friction  of  a  ball 
bearing  is  due  to  both  sliding  and  rolling  friction,  the 
sliding  friction  being  due  to  the  elasticity  of  the  balls 
and  the  bearing.  (See  Art.  159.)  Rolling  friction  is  most 


FRICTION  307 

nearly  approached  when  the  balls  are  hard  and  not  easily 
changed  from  their  spherical  shape.  All  materials,  how- 
ever, are  deformed  under  pressure  so  that  perfect  rolling 
friction  is  impossible.  On  account  of  the  sliding  friction 
present  in  roller  and  ball  bearings,  it  is  necessary  to  use  a 
lubricant  to  prevent  wear. 

Problem  383.  How  many  f-in.  balls  will  be  necessary  in  a  ball 
bearing  designed  to  carry  4000  lb.,  if  c  =  7.5?  If  /=  .0015,  what 
work  is  lost  per  revolution,  the  distance  from  the  axis  of  rotation  to 
the  center  of  balls  being  one  inch  ? 

164.  Friction  Gears.  —  In  the  friction  gears  the  driver 
is  usually  the  smaller  wheel,  and  when  there  is  any  differ- 
ence in  the  materials  of  which  the  wheels  are  made,  the 
driver  is  made  of  the  softer  material.  This  latter  arrange- 
ment is  resorted  to,  to  prevent  flat  places  being  worn  on 
either  wheel  in  case  of  slipping.  These  gears  have  been 
used  for  transmitting  light  loads  at  high  speeds,  where 
toothed  gears  would  be  very  noisy,  or  in  cases  where  it  is 
necessary  to  change  the  speed  or  direction  of  the  motion 
quickly. 

The  use  of  paper  drivers  has  made  possible  the  trans- 
mission of  much  heavier  loads  by  means  of  such  gears. 
A  series  of  tests,  made  by  W.  F.  M.  Goss,  and  reported 
in  Trans.  Am.  Soc.  M.  E.,  Vol.  18,  on  the  friction  be- 
tween paper  drivers  and  cast-iron  followers,  is  of  interest 
in  this  connection.  The  apparatus  used  is  shown  in  Fig. 
240.  The  pressure  between  the  wheels  was  obtained  by 
a  mechanism  that  forced  the  two  wheels  together  with  a 
pressure  P.  A  brake  wheel  shown  in  the  figure  absorbed 
the  power  transmitted. 


308 


APPLIED  MECHANICS  FOR  ENGINEERS 


The  coefficient  of  friction  was  regarded  as  the  ratio  of 
_Fto  .P,  as  in  sliding  friction.  While  this  is  customary, 
it  is  not  entirely  true,  since  we  have  the  rolling  of  one 


IRON  FOLLOWER 


ir 


SPRING 

[BALANCE 


FIG.  240 

body  over  the  other.     We  shall,  however,  assume  that  we 

F 

may  call  the  coefficient  of  friction  /=  —  •     It  was  found 

that  the  coefficient  of  friction  varied  with  the  slippage, 
but  was  fairly  constant  for  all  pressures  up  to  some  point 
between  150  to  200  Ib.  per  inch  of  width  of  wheel  face. 
"  Variations  in  the  peripheral  speed  between  400  and  2800 
ft.  per  minute  do  not  affect  the  coefficient  of  friction" 

If  the  allowable  coefficient  of  friction  be  taken  as  .20, 
the  horse  power  transmitted  per  inch  of  width  of  face  of 
the  wheel,  for  a  pressure  of  150  Ib.,  is 


H.P.  = 


150  x  .2  x  ^  nrd  x  w  x  N 
33,000 


FRICTION 


309 


where  d  is  the  diameter  of  the  friction  wheel  in  inches, 
w  the  width  of  its  face  in  inches,  and  iVthe  number  of 
revolutions  per  minute.  Using  this  formula,  the  following 
table  is  given  in  the  article  in  question: 

HORSE  POWER  WHICH  MAY  BE  TRANSMITTED  BY  MEAXS  OF  PAPER 

FRICTION  WHEEL  OF  ONE  INCH  FACE,  WHEN  RUN 

UNDER  A  PRESSURE  OF  150  LB. 


DIAMETER 
OP  PULLEY 
IN  INCHES 

REVOLUTIONS  PER  MINUTE 

25 

50 

75 

100 

150 

200 

60O 

100O 

8 

.0476 

.0952 

.1428 

.1904 

.2856 

.3808 

1.1424 

1.904 

10 

.0595 

.1190 

.1785 

.2380 

.3570 

.4760 

1.4280 

2.380 

14 

.0833 

.1666 

.2499 

.3332 

.4998 

.6664 

1.9992 

3.332 

16 

.0952 

.1904 

.2856 

.3808 

.5712 

.7616 

2.2848 

3.808 

18 

.1071 

.2142 

.3213 

.4284 

.6426 

.8568 

2.5704 

4.288 

24 

.1428 

.2856 

.4284 

.5712 

.8568 

1.1424 

3.4272 

5.712 

30 

.1785 

.3570 

.5355 

.7140 

1.0710 

1.4280 

4.2840 

7.140 

36 

.2142 

.4284 

.6426 

.8568 

1.2852 

1.7136 

5.1408 

8.560 

42 

.2499 

.4998 

.7497 

.9996 

1.4994 

1.9992 

5.9976 

9.996 

48 

.2856 

.5712 

.8568 

1.1424 

1.7136 

2.2848 

6.8544 

11.420 

The  value  of  the  coefficient  of  friction  for  friction 
gears  (Kent,  "  Pocket- 
Book ")  may  be  taken 
from  .15  to  .20  for 
metal  on  metal  ;  .25  to 
.30  for  wood  on  metal  ; 
.20  for  wood  on  com- 
pressed paper. 

Problem    384.      If    the 

friction  wheels  are  grooved 

as  shown  in  Fig.  241,  both  FIG.  241 


310 


APPLIED  MECHANICS  FOR  ENGINEERS 


of  cast  iron,  and  the  small  driver  fits  into  the  groove  of  the  larger 
follower,  prove  that  the  force  transmitted  is 


sm  a 

Problem  385.  The  speed  of  the  rim  of  two  grooved  friction 
wheels  is  400  ft.  per  minute.  If  a  =  45°,  /=  .18,  what  must  be  the 
pressure  P  to  transmit  100  horse  power? 

Problem  386.  What  horse  power  may  be  transmitted  by  the  gear- 
ing in  the  preceding  problem,  if  P  =  6000  Ib.  and  the  peripheral 
velocity  is  12  ft.  per  second  ? 


165.  Friction  of  Belts.  —  When  a  belt  or  cord  passes 
over  a  pulley  and  is  acted  upon  by  tensions  T±  and  Tv  the 
tensions  are  unequal,  due  to  the  friction  of  the  pulley  on 
the  belt.  We  shall  determine  the  relation  between  T±  and 
Ty  Let  the  pulley  be  represented  in  Fig.  242.  The  belt 


FIG.  242 


covers  an  arc  of  the  pulley  whose  angle  is  a.  Consider 
the  forces  acting  upon  the  belt  and  suppose  T^  and  Tz  to 
be  the  tensions  in  the  belt  on  the  tight  and  slack  sides, 


FRICTION 


311 


respectively,  and  ^the  tension  in  the  belt  at  any  point  of 
the  arc  of  contact.  Consider  the  forces  acting  on  a  por- 
tion of  the  belt  of  length  As  =  rA/3  (Fig.  242).  These 
forces  are  T,  T  +  A  T,  and  AP,  tangent  to  the  arc,  and  AP, 
the  normal  pressure  of  the  pulley  on  the  belt,  which  may 
be  regarded  as  acting  at  the  center  of  the  arc.  Let  m  be 
.the  mass  of  a  unit  length  of  the  belt.  The  mass  of  the 
portion  considered  is  then  wrA/3. 

Suppose  the  belt  to  have  uniform  speed,  v.     The  forces 
along  the  tangent  will  then  balance,  i.e. 

T+AF=  T+&T. 
.-.  dF=dT. 

y2 

The  acceleration  toward  the  center  is  — ,  and  the  force 
toward  the  center  is 

(T+  AT+  T)  sin^-AP. 
2 

.  •.  (2  T+  A27)  sin^  -  AP  =  rat>2A/3. 

2t 

Dividing  by  AyS  and  passing  to  the  limit,  remembering 


that 


lim 


•»¥ 


=  1,  we  have, 


If  /  is  the  coefficient  of  friction,  and  the  belt  is  on  the 
point  of  slipping, 

df-fdp, 


or 


312  APPLIED  MECHANICS  FOR  ENGINEERS 

Substituting  this  value  of  t?P,  we  have 


•  rp_  mvz 
Integrating, 


dT    =/a* 


T! 

=//3 

T2 


—  mv 


2 

.   Ti-mv2  _/*. 

'  T2- 


For  low  velocities  the  term  mv2  is  small  compared  to 
and  Ty  and  the  formula  may  be  written 


It  should  be  noted  that  m  in  the  above  formula  is  the 
mass  of  a  portion  of  the  belt  1  foot  long,  and  that  v  must 
be  reckoned  in  feet  per  second  if  g  is  taken  as  32.2. 

Problem  387.  A  rope  makes  two  complete  turns  around  a  post 
6  in.  in  diameter.  What  maximum  tension  could  be  balanced  by  a 
force  of  100  Ib.  if/=  .3? 

Problem  388.  A  weight  of  500  lb.  is  to  be  lowered  by  a  rope 
wound  round  a  horizontal  drum.  If  the  arc  of  contact  is  450°  and  the 
coefficient  of  friction  is  .25,  what  force  is  necessary  to  lower  the  weight 
uniformly  ? 

Problem  389.  The  velocity  of  a  belt  is  3000  ft./min.,  the  tension 
in  the  tight  side  is  150  lb.  per  inch  of  width,  the  coefficient  of  friction 
is  .25,  and  the  weight  of  a  portion  of  the  belt  1  ft.  long  and  1  in. 
wide  is  .15  lb.  If  the  belt  is  on  the  point  of  slipping  and  the  arc  of 
contact  is  150°,  what  is  the  tension  in  the  slack  side? 


FRICTION  313 

Problem  390.  A  rope  is  wrapped  four  times  around  a  post  and  a 
man  exerts  a  pull  of  50  Ib.  on  one  end.  If  the  coefficient  of  friction  is 
.3,  what  force  can  be  exerted  upon  a  boat  attached  to  the  other  end 
of  the  rope  V 

166.  Power  Transmitted  by  a  Belt.  —  From  the  relation 
/ 

dF=dT  (Art.  165.) 

there  follows 

/•T.  /*F 

I     dT=  \    dF, 

JTZ  J0 

or  2\  -  T2  =  F,  the  total  friction. 

The  work  done  per  second  against  F  is 

Fv  or  (  7\  -  TZ)V. 
Hence  the  horse  power  transmitted  by  the  belt  is 


where  2\  and  T2  are  in  pounds  and  v  is  in  feet  per  second. 

Problem  391.     Show  that  the  formula  for  H.P.  transmitted  by  the 
belt  may  be  reduced  to  the  form 

_(l 


550 

Problem  392.  Given  a  maximum  allowable  tension,  Tv  show  that 
the  power  transmitted  is  a  maximum  when 

.   •=>!£• 

Problem  393.  A  pulley  4  ft.  in  diameter  making  200  r.p.  m. 
drives  a  belt  that  absorbs  20  H.P.  The  belt  is  £  in.  thick  and  weighs 
56  Ib.  per  cubic  foot.  If  a  =  TT  audf  =  .27,  how  wide  must  the  belt  be 
that  the  tension  may  not  exceed  75  Ib.  per  inch  of  width  ? 


314 


APPLIED  MECHANICS  FOB  ENGINEERS 


Problem  394.  What  H.  P.  may  be  transmitted  by  a  belt  6  in. 
wide,  J  in.  thick,  weighing  56  Ib.  per  cubic  foot,  when  traveling  at 
1500ft.  per  minute,  if  a  =  108°,  f=  .25,  and  the  maximum  tension  is 
300  Ib.  per  square  inch  ? 

Problem  395.  Find  the  maximum  H.  P.  that  can  be  transmitted 
by  the  belt  in  the  preceding  problem,  and  the  corresponding  velocity. 

167.  Transmission  Dynamometer.  —  It  has  been  shown, 
in  Art.  165,  that  the  tension  of  a 
belt  on  the  tight  side  is  greater 
than  the  tension  on  the  slack  side. 
The  transmission  dynamometer 
(the  Fronde  dynamometer),  illus- 
trated in  Fig.  243,  is  designed  to 
measure  the  difference  in  these 
tensions.  Let  the  pulley  D  be  the 
driver  and  the  pulley  E  the  fol- 
lower, so  that  2j  represents  the 

*  |  tight  side  of  the  belt  and  T2  the 

slack  side.  The  pulleys  B,  B  run 
loose  on  the  T-shaped  frame  CBB. 
This  frame  is  pivoted  at  A.  If  we 
neglect  the  friction  due  to  the  loose 
pulleys,  we  have  the  following 
forces  acting  on  the  T-frame,  two 
forces  TI  at  the  center  of  the  right- 
hand  pulley  B,  two  forces  T2  at 
the  center  of  the  left-hand  pulley 
B,  a  measurable  reaction  P  at  (7, 

and  the  reaction  of  the  pin  at  A.     Taking  moments  about 

the  pin,  we  have 


FIG.  243 


FRICTION  315 

-  2 


so  that 


The  distances  CA  and  _&4.  are  known,  and  P  may  be 
measured  ;  the  difference,  then,  T^  —  Tv  may  always  be 
obtained.  The  value  T^  —  Tz  is  then  known  and  the  horse 
power  determined  by  the  relation 


33,000 

where  n  is  the  number  of  revolutions  per  minute,  and  r  is 
the  radius  of  the  machine  pulley  in  feet. 

168.  Creeping  or  Slip  of  Belts.  —  A  belt  that  transmits 
power  between  two  pulleys  is  tighter  on  the  driving  side 
than  it  is  on  ti\Q  following  side.  On  account  of  this  differ- 
ence in  tension  and  the  elasticity  of  the  material,  the  tight 
side  is  stretched  more  than  the  slack  side.  To  compen- 
sate for  this  greater  stretch  on  one  side  than  on  the  other, 
the  belt  creeps  or  slips  over  the  pulleys.  This  slip  has 
been  found  for  ordinary  conditions  to  vary  from  3  to  12 
ft.  per  minute.  The  coefficient  of  friction  when  the  slip 
is  considered  is  about  .27  (Lanza).  It  has  also  been 
found  that  the  loss  in  horse  power  in  well-designed  belt 
drives,  due  to  slip,  does  not  exceed  3  or  4  per  cent  of  the 
gross  power  transmitted,  and  that  ropes  are  practically 
as  efficient  as  belts  in  this  respect.  For  an  account  of  the 
experimental  investigations  on  this  subject  the  student  is 
referred  to  Inst.  Mech.  Eng.,  1895,  Vols.  3-4,  p.  599,  and 
Trans.  Am.  Soc.  M.  E.,  Vol.  26,  1905,  p.  584. 


316 


APPLIED  MECHANICS  FOR  ENGINEERS 


169.  Coefficient  of  Friction  of  Belting.  —  The  value  of  the 
coefficient  of  friction  of  belting  depends  not  only  on  the 
slip  but  also  upon  the  condition  and  material  of  the  rubbing 
surfaces.     Morin  found  for  leather  belts  on  iron  pulleys  the 
coefficient  of  friction  /=  .56  when  dry,  .36  when  wet,  .23 
when  greasy,  and  .15  when  oily  (Kent,  "Pocket-Book"). 
Most  investigators,  however,  including  Morin,  took  no  ac- 
count of  slip,  so  that  the  best  value  of  /,  everything  con- 
sidered, is  that  given  in  the  preceding  article  (.27). 

170.  Friction  of  a  Worn  Bearing.  —  The  friction  of  a 
bearing  that  fits  perfectly  is  the  friction  of  one  surface 
sliding  over  another  and  is  given  by  the  equation 

F=fN9 

where  F  is  the  force  of  friction,/  the  coefficient  of  friction, 
and  JVis  the  total  normal  pressure  on  the  bearing. 

When,  however,  the  hearing  is  worn,  as  is  shown  much 
exaggerated  in  Fig.  244,  the  friction  may  be  somewhat 

B 


FIG.  244 


FRICTION  317 

different.  When  motion  begins,  the  shaft  will  roll  up  on 
the  bearing  until  it  reaches  a  point  A  where  slipping  be- 
gins. If  motion  continues,  slipping  will  continue  along  a 
line  of  contact  through  A.  Let  P  be  a  force  that  causes 
the  rotation,  R  a  force  tending  to  resist  the  rotation,  and 
El  the  reaction  of  the  bearing  on  the  shaft.  There  are 
only  three  forces  acting  on  the  shaft,  so  that  P,  R,  and  Rl 
must  meet  in  the  point  B.  The  direction  of  Rl  is  accord- 
ingly determined.  The  normal  pressure  is  N=  R1  cos  0, 
and  the  force  of  friction  is 


It  is  seen  that  6  is  the  angle  of  friction.     The  moment  of 
the  friction  with  respect  to  the  center  of  the  axle  is 

Fr  =  Rir  sin  0. 

If  the  axle  is  well  lubricated,  so  that  6  is  small  and  sin  6 
may  be  replaced  by  tan  6  =/,  the  friction  is 


and  the  moment 


The  circle  tangent  to  AS  of  radius  r  sin  0  is  called  the 
friction  circle.  Since  r  and  6  are  known  generally,  this 
circle  may  be  made  use  of  in  locating  the  point  A. 

The  shaft  will  continue  to  rotate  in  the  bearing  so  long 
as  the  reaction  Rl  falls  within  the  friction  circle,  and 
slipping  will  begin  as  soon  as  the  direction  of  R^  becomes 
tangent  to  the  friction  circle. 

Problem  396.  If  the  radius  of  the  siiaft  is  1  in.,  0  =  4°, 
P  =  500  lb.,  aj  =  3  ft.,  a2  =  2  ft.,  angle  between  ai  anda2  is  100°,  and 


318 


APPLIED  MECHANICS  FOR  ENGINEERS 


P  and  R  are  right  angles  to  al  and  a2,  what  resistance  R  may  be 
overcome  by  P  when  slipping  occurs  ? 

Problem  397.  The  radius  of  a  shaft  is  1  in.,  R  =  29.5  lb., 
P  =  20  lb.,  ax  =  3  ft.,  and  a2  =  2  ft.  What  force  of  friction  will  be 
acting  at  the  point  A,  when  the  angles  between  P  and  ax  and  R  and 

a2  are  right  angles  ?  What  must  be  the 
value  of  the  coefficient  of  friction?  The 
angle  between  ax  and  a2  is  110°. 

171.  Friction  of  Pivots.  —  The 
friction  of  pivots  presents  a  case  of 
sliding  friction,  so  that  the  force  of 
friction  F  equals  the  coefficient  of 
friction  times  the  normal  pressure. 

That  is, 

F  =  fN. 

(a)  Flat-  end  Pivot.  Assuming 
the  pivot  to  press  uniformly  on 
the  bearing,  the  friction  on  an 
element  of  area  r'dddr'  is 

fJLr<dOdr'.     (Fig.  245.) 


FIG.  245 


The  total  work  done  against  fric- 
tion  per  revolution  is 


Work  per  rev.  =  P  f^Trr'/— 

Jo    Jo  J  2 


or 


Work  per  re  volution  = 

Collar  Bearing  or  Hollow  Pivot 


FRICTION  319 

Here  the  work  per  revolution  becomes  (Fig.  246) 

Work  per  revolution  =  f  **2  (* ^  2  irr'    ,  f**  ax  r'dr'dQ 

Jri  Jo  ,r(rg  -  r») 


FIQ.  246 

(<?)  Conical  Pivot.  The  conical  pivots,  illustrated  in 
Fig.  247,  do  not  usually  fit  into  the  step  the  entire  depth 
of  the  cone.  Let  the  radius  of  the  cone  at  the  top  of  the 
step  be  rv  a  half  the  angle  of  the  cone,  and  let  dP1  be  the 
normal  pressure  of  the  bearing  on  an  elementary  area  whose 
horizontal  projection  is  r'dr'dO. 


320 


APPLIED   MECHANICS  FOB   ENGINEERS 


The  vertical  com- 
ponent of  dP1  is 
equal  to  the  vertical 
load  on  the  hori- 
zontal area  r'dr'dO, 
which  is 


dP  =  —r'dr'd0. 


dp  =  Pr'dr'd9 
irr\  sin  a 

and  the  force  of  fric- 
tion acting  on  this 
element  is 


FIG.  247 

The  total  work  per 
revolution  done  against  friction  is  therefore 


Work  per  revolution  =  f  **'  f  2" 

Jo    Jo    itr\ 


itr\  sin  a 


3   sin  a 


If  «  =  -,  this  value  for  work  lost  reduces  to  the  work 

4H 

lost  per  revolution,  in  the  case  of  the  flat-end  solid  pivot. 
It  is  easily  seen,  since  sin  a  is  less  than  unity,  that  if  r^  is 
nearly  equal  to  r,  the  friction  of  the  conical  bearing  is 
greater  than  the  friction  of  the  flat-end  bearing.  This 
might  have  been  expected  from  the  wedgelike  action  of 
the  pivot  on  the  step.  It  is  also  easily  seen  that  ^  may 


FRICTION 


321 


be  taken  small  enough  so  that  the  friction  will  be  less 
than  the  friction  of  the  flat  pivot.  The  work  lost  due  to 
friction  in  the  case  of  the  conical  pivot  will  be  equal  to, 
greater,  or  less  than,  the  work  lost,  due  to  friction  in  the 
case  of  the  flat-end  pivot, 
according  as  \p 

r+>r  sin  a. 
< 

(d)  Spherical  Pivot. 
Suppose  the  end  of  the 
pivot  is  a  spherical  sur- 
face, as  shown  in  Fig. 
248.  Let  r  be  the  radius 
of  the  shaft  and  r1  the 
radius  of  the  spherical 
surface;  then  the  load 

per    unit    of     area     of 

p 
horizontal  surface  is  — ^. 


o      r 


FIG.  248 


The  horizontal  projection  of  any  elementary  ring  of  the 
bearing,  of  radius  x,  is  2  irxdx.     The  load  on  this  area  is 

ZPxdx 


and  the  corresponding  normal  pressure  is 

2Pxdx 


sec 


and  the  total  work  done  against  friction  in  one  revolution 
is 


r- 


r2  cos 


322  APPLIED  MECHANICS  FOR   ENGINEERS 

Here  /3  is  a  variable  such  that 

x  =  r1  sin  /3. 
.-.  dx  —  r^  cos/3c?/3, 

and  the  expression  for  the  work  becomes 

Work  per  rev.  =  l^^i  f  "  sin2  j3d/3 
*     ^ 


since  r  =  r  sin  a. 


If   the  bearing  is  hemispherical,  a  =  -,  and  the  work 

2 
lost  per  revolution  becomes 


The  friction  of  flat  pivots  is  often  made  much  less  by 
forcing  oil  into  the  bearing,  so  that  the  shaft  runs  on  a 
film  of  oil.  In  the  case  of  the  turbine  shafts  of  the  Niag- 
ara Falls  Power  Company  (see  Art.  143)  the  downward 
pressure  is  counteracted  by  an  upward  water  pressure. 
In  some  cases  the  end  of  a  flat  pivot  has  been  floated  on 
a  mercury  bath.  This  reduces  the  friction  to  a  minimum. 
(See  Engineering,  July  4,  1893.) 

The  Schiele  pivot  is  a  pivot  designed  to  wear  uniformly 
all  over  its  surface.  The  surface  is  a  tractrix  of  revolu- 
tion ;  that  is,  the  surface  formed  by  revolving  a  tractrix 
about  its  asymptote.  Its  value  as  a  thrust  bearing  is 
not  as  great  as  was  first  anticipated.  (See  American  Ma- 
chinist, April  19,  1894.) 


FRICTION  323 

The  coefficient  of  friction  for  well-lubricated  bearings 
of  flat-end  pivots  has  been  found  to  vary  from  .0044  to 
.0221.  (See  Proc.  Inst.  M.  E.,  1891.)  For  poorly  lubri- 
cated bearings  the  coefficient  may  be  as  high  as  .10  or 
•.25  for  dry  bearings. 

Problem  398.  Show  that  the  work  lost  per  revolution  for  the 
hemispherical  pivot  is  2.35  times  the  work  lost  per  revolution  for  the 
flat  pivot. 

Problem  399.     The  entire  weight  of  the  shaft  and  rotating  parts 

•  of  the  turbines  of  the  Niagara  Falls  power  plant  is  152,000  lb.,  the 

diameter  of  the  shaft  11  in.     If  the  coefficient  of  friction  is  con-i 

sidered  as  .02  and  the  bearing  a  flat-end  pivot,  what  work  would  be 

lost  per  revolution  due  to  friction  ? 

Problem  400.  A  vertical  shaft  carrying  20  tons  revolves  at  a 
speed  of  50  revolutions  per  minute.  The  shaft  is  8  in.  in  diameter 
and  the  coefficient  of  friction,  considering  medium  lubrication,  is 
.08.  What  work  is  lost  per  revolution  if  the  pivot  is  flat?  What 
horse  power  is  lost  ?  What  horse  power  if  the  pivot  is  hemispherical  ? 

Problem  401.  What  horse  power  would  be  lost  if  the  shaft  in 
the  preceding  problem  was  provided  with  a  collar  bearing  18  in.  out- 
side diameter  instead  of  a  flat-end  block?  Compare  results. 

Problem  402.  A  vertical  shaft  making  200  revolutions  per  minute 
carries  a  load  of  20  tons.  The  shaft  is  6  in.  in  diameter  and  is 
provided  with  a  flat-end  bearing,  well  lubricated.  If  the  coefficient 
of  friction  is  .004,  what  horse  power  is  lost  due  to  friction? 

172.  Absorption  Dynamometer.  —  The  friction  brake  shown 
in  Fig.  240  is  used  to  absorb  the  energy  of  the  mechanism. 
It  may  be  used  as  a  means  of  measuring  the  energy,  and 
when  so  used  it  may  be  called  an  absorption  dynamometer. 
The  weight  IP,  attached  to  one  end  of  the  friction  band, 
corresponds  to  the  tension  in  the  tight  side  of  an  ordinary 


324  APPLIED  MECHANICS   FOR  ENGINEERS 

belt  (see  Art.  165),  while  the  force  measured  by  the  spring 
S  corresponds  to  the  tension  on  slack  side  of  a  belt.  Let 
TF=  T^  and  S=T^  then  T^  =  T^*,  just  as  was  found  in 
the  case  of  belt  tension.  The  work  absorbed  per  revolu- 
tion is  work  =  (T1—  T^  TTTV  where  rl  is  the  radius  of  the 
brake  wheel.  The  horse  power  absorbed  is 


HP  = 


33,000 


where  n  is  the  number  of  revolutions  per  minute. 
.    In  many  cases  the  friction  band  is  a  hemp  rope,  and  in 
such  cases  it  is  possible  to  wrap  the  rope  one  or  more  times 
around  the  pulley,  making  it  possible  to  make   2\  —  T2 
large  while  T2  is  small. 

The  surface  of  the  brake  wheel  may  be  kept  cool  by  allow- 
ing water  to  flow  over  the  inside  surface  of  the  rim,  which 
should  be  provided  with  inside  flanges  for  that  purpose. 

173.  Friction  Brake.  —  The  friction  brake  shown  in 
Fig.  249  consists  of  the  lever  EC,  the  friction  band,  and 
the  friction  wheel.  Such  brakes  are  used  on  many  types 
of  hoisting  drums,  automobiles,  etc.  Let  the  band  ten- 
sions be  jPj  and  jT2,  and  let  W  be  the  force  causing  the 
motion,  that  is,  the  working  force,  and  P  the  force  applied 
at  the  end  of  the  lever  EG  in  such  a  way  as  to  retard  the 
rotation  of  the  drum.  We  have  here  as  before  T±  =  T2e/a 
and  the  work  per  revolution  (  2\  —  T2)  2  7rr2  +  F"2  7rrB. 
By  taking  moments  about  A  AVC  have  for  uniform  motion 

Tj  -  Tz  =  Wri  ~  Fr*,  where  rs  is  the  radius  of  the  shaft 

T2 

and  F  is  the  force  of  friction  acting  on  the  shaft.     Taking 


FRICTION 


325 


moments  about  (7,  we  have 

=  Tldl  sin  8  + 


sn 


FIG.  249 

Problem  403.  A  weight  of  1  ton  is  lowered  with  uniform  velocity 
by  means  of  a  friction  brake  as  in  Fig.  249,  in  which  the  lower  por- 
tion of  the  belt  is  horizontal.  Given  radius  of  drum  =  1^  ft.,  r,adius 
of  friction  wheel  =  2  ft.,  coefficient  of  brake  friction  =  .30,  coefficient 
of  axle  friction  =  .04,  radius  of  shaft  =  1  in.,  8  =  45°,  /?  =  15°, 
di  =  dz  —  1  ft.,  EC  =  6  ft.,  and  weight  of  drum  and  brake  wheel 
=  600  lb.,  find  7\  and  T2,  and  P. 

(SUGGESTION  :  Find  771  and  Tv  neglecting'axle  friction ;  from  these 
values  obtain  axle  friction  and  recompute.) 

Problem  404.  What  force  P  will  be  necessary  to  reduce  the  ve- 
locity of  the  above  weight  from  Wf/s  to  5f/s  in  10  ft.,  if  the  radius 
of  gyration  of  drum  and  brake  wheel  is  1 J  ft.  ?  What  then  will  be 
the  values  of  Tv  T^  and  the  tension  in  the  rope  supporting  W? 

Problem  405.  If  the  weight  in  the  above  problem  has  a  velocity 
of  10  ft.  per  second,  and  it  is  required  that  the  mechanism  be  so  con- 
structed that  it  could  be  stopped  in  a  distance  of  6  ft.,  what  pressure 
P  on  the  lever  and  tensions  7\  and  T2  would  it  require?  What 


326 


APPLIED  MECHANICS  FOR  ENGINEERS 


would  be  the  tension  in  the  rope  caused  by  the  stop?    Compare  this 
tension  with  W,  the  tension  when  the  motion  is  uniform. 

174.  Prony  Friction  Brake.  —  The  Prony  friction  brake 
may  be  used  as  an  absorption  dynamometer  as  shown  in 
principle  in  Fig.  250.  Let  W  be  a  working  force  acting 
on  the  wheel  of  radius  r  and  suppose  the  brake  wheel  to 


FIG.  250 

be  of  radius  rr  The  brake  consists  of  a  series  of  blocks 
of  wood  attached  to  the  inner  side  of  a  metal  band  in 
such  a  way  that  it  may  be  tightened  around  the  brake 
wheel  as  desired  by  a  screw  at  B.  This  band  is  kept  from 
turning  by  a  lever  CAD,  held  in  the  position  shown  by  an 
upward  pressure  P,  at  A.  Considering  the  forces  acting 
on  the  brake  and  taking  moments  about  the  center,  we 
have  the  couple  due  to  friction,  Frv  equal  to  the  moment 


Considering  the  forces  acting  on  the  wheel,  and  neglecting 
axle  friction,  we  get  for  uniform  motion  of  W, 

F^  =  Wr. 


FRICTION 


327 


The  energy  absorbed  is  used  in  heating  the  brake  wheel. 
The  wheel  is  kept  cool  by  water  on  the  inside  of  the  rim. 
The  work  absorbed  per  minute  is  2  irr^Fn  =  2  7rP(  OA)n, 
where  n  is  the  number  of  revolutions  per  minute.  The 
force  P  may  be  measured  by  allowing  a  projection  of  the 
arm  at  A  to  press  upon  a  platform  scales.  The  horse 
power  absorbed  is 


33,000 

where  OA  is  expressed  in  feet,  and  n  is  the  number  of 
revolutions  per  minute. 

33 
If  OA  be  taken  as  -  —  ,  a  convenient  length,  the  formula 

'2  7T 

reduces  to 

„  TO        Fn 

=  Iooo' 

A  dynamometer  slightly  different  from  the  Prony  dy- 
namometer is  shown  in  Fig.  251.  It  differs  only  in  the 
means  of  measuring  P.  In  this  case  the  force  P  is  meas- 


FIG.  251 


328  APPLIED  MECHANICS  FOR  ENGINEERS 

ured  by  the  angular  displacement  of  a  heavy  pendulum 
WY  Taking  moments  about  the  axis  of  W±  and  calling 
r5  the  distance  from  that  axis  to  its  center  of  gravity  and 
y8  the  angular  displacement,  we  have 


so  that  the  horse  power  absorbed  may  be  written 

_  2  ir(  OA)  JFin  sin  pw, 
=  r4  33,000 

where  OA,  r4,  and  r&  are    expressed  in  feet.     If  OA  be 

33 
taken  as  —  ,  this  becomes 

27T? 

w  T»  _  W\r*  sin  fin 

~~ 


The  student  should  understand  that  the  rotation  of  the 
mechanism  at  0  is  not  in  every  case  due  to  a  weight  W 
being  acted  upon  by  gravity.  In  fact,  in  most  cases,  the 
motion  will  be  due  to  the  action  of  some  kind  of  engine. 
This,  however,  will  not  change  the  expressions  for  horse 
power. 

OO 

Problem  406.   If    Wl  =  100  lb.,  OA  =  —-,  rs  =  2  ft.,  and  r4  =  6 

in.,  what  horse  power  is  absorbed  by  the  brake  if  ft  is  30r',  and  n  is 
300  revolutions  per  minute  ? 

175.  Friction  of  Brake  Shoes.  —  The  application  of  the 
brake  shoe  to  the  wheel  of  an  ordinary  railway  car  is 
shown  in  Fig.  252,  where  F'  is  the  axle  friction,  F  the 
brake-shoe  friction,  N  the  normal  pressure  of  the  brake 
shoe,  G-  the  weight  on  the  axle,  and  F^  and  N-±  the  reaction 


FRICTION 

of  the  rail  on  the  wheel.  The  brakes  on  a  railway  car 
when  applied  should  be  capable  of  absorbing  all  the  en- 
ergy of  the  car  in  a 
very  short  time.  The 
high  speeds  of  modern 
trains  require  a  system 
of  perfectly  working 
brakes,  capable  of  stop- 
ping the  car  when 
running  at  its  maxi- 
mum speed  in  a  very  _ 

short  distance.  '          •  ' 

FIG.  252 

The     coefficient     of 

friction  between  the  shoes  and  wheel  for  cast-iron  wheels 
at  a  speed  of  40  mi.  per  hour  is  about  ^,  while  at  a  point 
15  ft.  from  stopping  the  coefficient  of  friction  is  increased 
7  per  cent,  or  it  is  about  .27.  The  coefficient  for  steel- 
tired  wheels  at  a  speed  of  65  mi.  per  hour  is  .15,  and  at  a 
point  15  ft.  from  stopping  it  is  .10.  (See  Proc.  M.  C.  B. 
Assoc.,  Vol.  39,  1905,  p.  431.) 

The  brake  shoes  act  most  efficiently  when  the  force  of 
friction  F  is  as  large  as  it  can  be  made  without  causing  a 
slipping  of  the  wheel  on  the  rail  (skidding).  The  normal 
pressure  JVJ  corresponding  to  the  values  of  the  coefficient 
of  friction  given  above,  varies  in  brake-shoe  tests  from 
2800  Ib.  to  6800  lb.,  sometimes  being  as  high  as  10,000  Ib. 

Problem  407.  A  20- ton  car  moving  on  a  level  track  with  a 
velocity  of  a  mile  a  minute  is  subjected  to  a  normal  brake-shoe  pres- 
sure of  6000  lb.  on  each  of  the  8  wheels.  If  the  coefficient  of  brake 
friction  is  .15,  how  far  will  the  car  move  before  coming  to  rest? 


330  APPLIED  MECHANICS  FOB  ENGINEERS 

Problem  408.  In  the  above  problem  the  kinetic  energy  of  rota- 
tion of  the  wheels,  the  axle  friction,  and  the  rolling  friction  have  been 
neglected.  The  coefficient  of  friction  for  the  journals  is  .002,  that 
for  rolling  friction  is  .02.  Each  pair  of  wheels  and  axle  has  a  mass 
of  45  and  a  moment  of  inertia  with  respect  to  the  axis  of  rotation  of 
37.  The  diameter  of  the  wheels  is  32  in.  and  the  radius  of  the  axles 
is  2£  in.  Compute  the  distance  the  car  in  the  preceding  problem  will 
go  before  coming  to  rest.  Compare  the  results. 

Problem  409.  A  30-ton  car  is  running  at  the  rate  of  70  mi.  per 
hour  on  a  level  track  when  the  power  is  turned  off  and  brakes  ap- 
plied so  that  the  wheels  are  just  about  to  slip  on  the  rails.  If  the 
coefficient  of  friction  of  rest  between  wheels  and  rails  is  .20,  how 
far  will  the  car  go  before  coming  to  rest  ? 

Problem  410.  A  75-ton  locomotive  going  at  the  rate  of  50  mi. 
per  hour  is  to  be  stopped  by  brake  friction  within  2000  ft.  If  the 
coefficient  of  friction  is  .25,  what  must  be  the  normal  brake-shoe 
pressure  ? 

Problem  411.  A  75-ton  locomotive  has  its  entire  weight  carried 
by  five  pairs  of  drivers  (radius  3  ft.).  The  mass  of  one  pair  of  drivers 
is  271  and  the  moment  of  inertia  is  1830.  If,  when  moving  with  a 
velocity  of  50  mi.  per  hour,  brakes  are  applied  so  that  slipping  on 
the  rails  is  impending,  how  far  will  it  go  before  being  stopped  ?  The 
coefficient  of  friction  between  the  wheels  and  rails  is  .20. 

176.  Train  Resistance.  —  The  resistance  offered  by  a  train 
depends  upon  a  number  of  conditions,  such  as  velocity, 
acceleration,  the  condition  of  track,  number  of  cars,  curves, 
resistance  of  the  air,  and  grades.  No  law  of  resistance 
can  be  worked  out  from  a  theoretical  consideration,  be- 
cause of  the  uncertainty  of  the  influence  of  the  various 
factors  involved.  Formulae  have  been  developed  from  the 
results  of  tests ;  the  most  important  of  these  are  given 
below. 


FRICTION  331 

Let  R  represent  the  resistance  in  pounds  and  v  the 
velocity  in  miles  per  hour.  W.  F.  M.  Goss  has  found  that 
the  resistance  may  be  expressed  as 


where  L  is  the  length  of  the  train  in  feet.  (See  Engineer- 
ing Record,  May  25,  1907.) 

The    Baldwin   Locomotive    Works    have    derived    the 

formula  R  =  3  +  ^ 

o 

as  the  relation  between  the  resistance  and  velocity.  When 
all  factors  are  considered,  this  becomes 

R  =  3  +  -  +  .3788(0  +  .5682(<?)  +  .1265(a), 
6 

where  t  =  grade  in  feet  per  mile,  c  the  degree  of  curvature 
of  the  track,  and  a  the  rate  of  increase  of  speed  in  miles 
per  hour  in  a  run  of  one  mile. 

To  get  the  total  resistance  it  is  necessary  to  include,  in 
addition  to  the  above  factors,  the  friction  of  the  locomo- 
tive and  tender.  This  is  given  by  Holmes  (see  Kent's 
"Pocket-Book")  as 


where  Wis  the  weight  of  the  engine  and  tender  in  pounds 
and  Rl  the  resistance  in  pounds  due  to  friction. 

Other  formulae  derived  as  the  result  of  experiments  are 
shown  graphically  in  Fig.  253. 


332  APPLIED  MECHANICS  FOR  ENGINEERS 


RESISTANCE  VELOCITY  CURVES 
FOR  RAILROAD  TRAINS 


40         50        60        70        80         90       100 
VELOCITY  IN  MILES  PER  HOUR 

FIG.  253 


FRICTION  ' 


333 


The  formulae  themselves  are  as  follows  (see  Engineering, 
July  26,  1907)  : 


CURVE  NUMBER 

FORMULA 

AUTHORITY 

1 

R  =  12  +  is 

Clark 

2 

R  =  8  +  ^i 

Clark 

3 

*  =  4.48  +  I| 

Wellington 

4 

v2 

Deeley 

5 
6 

7 
8 

9 

.R  =  .2497  v 
R  =  3.36  +  .1867  v 
R  =  4.48  +  284  v 
R  =  2  +  .24  v 

Laboriette 
Baldwin  Company 
Lundie 
Sinclair 

Aspinal 

R            +  65.82 

It  is  evident  that  these  formulae  do  not  agree  as  closely 
as  one  would  wish.  The  difference  must  be  due  chiefly  to 
the  different  conditions  under  which  the  tests  were  made. 
These  conditions  should  be  taken  into  account  in  any 
application  of  the  formulae  to  special  cases. 


CHAPTER  XIV 
DYNAMICS  OF  RIGID  BODIES 

177.  Statement  of  D'Alembert's  Principle.  — A  body  may 
be  considered  as  made  up  of  a  collection  of  individual 
particles  held  together  by  forces  acting  between  them. 
The  motion  of  a  body  concerns  the  motion  of  its  individ- 
ual particles.  We  have  seen  that  in  dealing  with  such 
problems  as  the  motion  of  a  pendulum  it  was  necessary  to 
consider  the  body  as  concentrated  at  its  center  of  gravity ; 
that  is,  to  consider  it  as  a  material  point.  The  principle 
due  to  D'Alembert  makes  the  consideration  of  the  motion 
of  bodies  an  easy  matter.  Consider  a  body  in  motion  due 
to  the  application  of  certain  external  forces  or  impressed 
forces.  Instead  of  thinking  of  the  motion  as  being  pro- 
duced by  such  impressed  forces,  imagine  the  body  divided 
into  its  individual  particles  and  imagine  each  of  the  parti- 
cles acted  upon  by  such  a  force  as  would  give  it  the  same 
motion  it  has  due  to  the  impressed  forces.  These  forces 
acting  upon  the  individual  particles  are  called  the  effective 
forces.  D'Alembert's  Principle,  then,  states  that  the  im- 
pressed forces  will  be  in  equilibrium  with  the  reversed  effec- 
tive forces. 

This  amounts  to  saying  that  the  system  of  effective 
forces  is  equivalent  to  the  system  of  impressed  forces. 
The  truth  of  the  principle  is  evident  if  it  be  granted  that 

334 


DYNAMICS   OF  EIGID  BODIES 


335 


the  forces  acting  between  any  two  particles  of  the  body 
are  equal  and  opposite  and  act  in  the  line  joining  the 
particles.  In  an^  summation  of  all  the  forces  acting  on 
all  of  the  particles  of  the  body  these  forces  acting  between 
the  particles  of  the  body  would  cancel  out  and  leave  the 
summation  of  the  impressed  forces. 

178.  Simple  Translation  of  a  Rigid  Body.  —  A  body  has 
pure  translation  when  all  points  of  the  body  have  at  each 
instant  the  same  velocity. 

Consider  a  body  (Fig. 
254)  under  the  action  of 
the  impressed  forces  Pv  P2, 
P3,.-.,  and  let  a  be  the  ac- 
celeration of  all  particles  of 
the  body.  Choose  axes  of 
reference,  taking  the  #-axis 
parallel  to  the  direction  of 
the  acceleration.  Think  of 
the  body  as  divided  up  into 
particles  and  each  particle 
under  the  action  of  the  effective  forces.  Let  dM  be  the 
mass  of  any  particle  of  the  body.  The  effective  forces 
acting  on  dM  can  be  combined  into  a  single  force,  dF, 
parallel  to  the  direction  of  the  acceleration,  according  to 
Newton's  second  law,  such  that 
dF=a  -dM. 

Since  the  effective  forces  are  parallel  and  proportional 
to  the  weights  of  the  elements  of  mass  on  which  they  act, 
their  resultant  is  a  force,  F,  parallel  to  the  direction  of  the 


FIG.  254 


336 


APPLIED  MECHANICS  FOR  ENGINEERS 


acceleration,  through  the  center  of  gravity  of  the  body,  such 
that 


or 


F=Ma. 


Hence  the  resultant  of  the  impressed  forces  must  be  a 
force,  M  •  a,  parallel  to  the  direction  of  the  acceleration  and 
acting  through  the  center  of  gravity. 

Therefore,  if  a  body  has  a  motion  of  pure  translation, 
the  resultant  of  all  impressed  forces  acting  on  the  body  is 
parallel  to  the  direction  of  the  acceleration  and  acts 

through  the  center 
of  gravity  of  the 
body.  Also,  the 
acceleration  of  the 
center  of  gravity 
of  the  body  is  the 
same  as  if  the 
whole  mass  were 
concentrated  at  the 
center  of  gravity 
and  the  resultant 
force  acted  at  that 
point. 


FIG.    255 


179.  Rotation  of 
a  Body  about  a 
Fixed  Axis.  —  Let 
Fig.  255  represent  a  body  rotating  about  a  fixed  axis,  here 
chosen  as  the  axis  of  z,  under  the  action  of  forces,  f  P 


DYNAMICS  OF  BIGID  BODIES  337 

P8,  ....  Let  a  be  the  angular  acceleration  of  the  body  at 
any  instant.  Think  of  the  body  as  divided  up  into  infini- 
tesimal elements  and  let  dM  be  the  mass  of  any  such 
element,  distant  r  from  the  axis  of  rotation.  The  effec- 
tive forces  acting  on  this  element  can  be  combined  into 
two  components,  dT  and  dN,  respectively,  along  the 
tangent  and  normal  to  the  circular  path,  since  the  motion 
of  the  particle  is  in  a  circle  whose  plane  is  perpendicular 
to  the  axis  of  rotation.  The  values  of  dT  and  dZVare 

dT  =  atdM  =  radM, 

dN  =  rMM.  (Arts.  117,  126.) 


Applying  D'Alembert's  principle,  and  taking  moments 
of  reversed  effective  forces  and  impressed  forces  about 
OZ,  we  get 

2  moms  impressed  forces  —  irdT  =  0, 

or  2  mom  —  Cr2adM=  0, 

or  S  mom  =  a  J, 

in  which  I  is  the  moment  of  inertia  of  the  body  with  re- 
spect to  the  axis  of  rotation. 

Therefore,  when  a  body  rotates  about  a  fixed  axis,  the 
sum  of  the  moments  of  the  impressed  forces  about  the 
axis  of  rotation  is  equal  to  the  angular  acceleration  of 
the  body  times  the  moment  of  inertia  of  the  body  about 
the  axis  of  rotation. 

180.  Equations  of  Motion  of  the  Rotating  Body.  —  In  addi- 
tion to  the  above  equation  there  are  five  other  equations 
which  express  the  conditions  of  equilibrium  of  the  im- 


338  APPLIED  MECHANICS  FOR  ENGINEERS 

pressed  forces  and  the  reversed  effective  forces  (Art.  177). 

The  six  conditions  may  be  stated  as  follows  :  The  sum 

of  the  components  of  the 
impressed  forces  parallel  to 
each  of  the  three  axes  equals 
the  sum  of  the  components 
of  the  effective  forces  par- 
allel to  those  axes,  and  the 
sum  of  the  moments  of  the 
impressed  forces  about  each 
of  the  axes  equals  the  sum 
of  the  moments  of  the  effec- 

FIG.  256 

tive  forces  about  those  axes. 

Representing  impressed  forces  with  a  subscript,  i,  and 
reckoning  moments  in  the  direction  shown  in  Fig.  256, 
these  six  conditions  may  be  written, 


=  -  J*cos  0  dN-  J"sin 

2  Yi  =  -  J"sin  <j>dN+Cc 
2^=0; 
2  momia.  =  j  z  sin  (f>dN—  \  z  cos  <f>dT, 

2  mom^  =  —  \  z  cos  (f>dN—  \  z  sin 
IS  momi2  =  j  rdT. 

Replacing  dN  and  dT  by  their  values  from  Art.  179, 
and  replacing  r  cos  0  and  r  sin  <£  respectively  by  x  and  ^, 
these  equations  reduce  to  the  forms, 


DYNAMICS  OF  RIGID  BODIES  339 


(1) 

2  F;  =  -  o)2pf  +  axM,  (2) 

2^=0;  (3) 

2  momfa  =  w2  j  yzdM  —  a  \  xzdM,  (4) 

2  mom^  =  —  w2  CxtdM—  a  \  yzdM,  (5) 

2  mom*,  =  alz.  (6) 

These  equations  hold  true  at  any  instant  of  the  motion  of 
the  body.  Since  x  and  y  are  the  coordinates  of  the  center 
of  gravity,  if  the  axis  of  rotation  passes  through  the 
center  of  gravity,  the  right-hand  sides  of  equations  (1) 
and  (2)  reduce  to  zero. 

Also  if  the  ir^-plane  is  a  plane  of  symmetry  of  the  body, 

J  xzdM  and   J  yzdM  both  reduce  to  zero,  since  for  every 

term  #(+  z)dM  there  is  a  corresponding  term  #(—  z)dM^ 
and  for  every  term  y(+z)dM  there  is  a  corresponding 
term  ^(—  z)dM. 

Therefore,  when  the  axis  of  rotation  passes  through 
the  center  of  gravity,  and  is  taken  as  the  z-axis,  and  the 
a^y-plane  is  a  plane  of  symmetry  of  the  body,  the  six 
equations  for  the  forces  acting  on  the  body  become  : 

2^=0,  (!') 

(20 
(30 

2momfe=0,  (4') 

2  momiy  =  0,  (5') 

2 


340 


APPLIED  MECHANICS  FOR   ENGINEERS 


This  case  is  the  one  that  usually  comes  up  in  engineer- 
ing problems,  and  so  these  simplified  equations  are  more 
often  used  than  the  six  more  general  equations.  It  will 
be  noticed  that  these  equations  are  exactly  the  same  as 
the  conditions  for  equilibrium  as  determined  in  Art.  56, 
except  that  2  momiz  is  not  zero. 

181.  Reaction  of  Supports  of  a  Rotating  Body.  —  When 
a  body  is  rotating  about  a  fixed  axis,  the  reactions  of  the 
supports  on  the  body  are  impressed  forces  and  the  appli- 
cations of  equations  (1)  to  (6)  of  the  preceding  article  will, 

when  the  other  impressed 
forces  and  the  angular  veloc- 
ity are  given,  enable  one  to 
find  the  reactions  of  the  sup- 
ports. As  an  illustration 
suppose  the  body  (Fig.  257) 
to  be  a  cast-iron  sphere,  radius 
2  in.,  connected  to  a  vertical 
axis  by  a  weightless  arm 
whose  length  is  4  in.,  the 
axis  being  supported  at  two 
points  as  shown.  The  reac- 
tions of  the  lower  support 
are  Px,  Py,  Pz,  and  of  the 
upper  PI,  PI,  0.  Let  the 
body  be  rotated  by  a  cord 
running  over  a  pulley  of 
radius  1  in.  situated  2  in. 
below  PI.  Let  the  constant  tension,  P,  in  the  cord  be  10 
Ib.  and  suppose  it  acts  parallel  to  the  «/-axis.  Consider 


FIG.  257 


DYNAMICS  OF  RIGID  BODIES  341 

the  motion  when  the  sphere  is  in  the  22-plane.     Take  the 
#?/-plane  through  the  center  of  the  sphere  perpendicular 

to  2,  then   (xzdM'dud  I  yz dM  are  both  zero.     Using  the 

foot-pound-second  system  of  units,  and  the  weight  of  cast 
iron   as   450   Ib.    per   cubic  foot,  we  have  x  =  J,  y  =  0, 
a  =  8.727  Ib.,  M=  .271,  and  I2  =  .0708. 
Equations  (1)  to  (6)  (Art.  180)  then  become, 

PX  +  P<X  =  -.  1355  a,',  (1) 

P,  +  P;-10  =      .1355«,  (2) 

Pz-  8. 727  =  0,  (3) 

¥-A'  +  i^  =  0,  (4) 

P^-JP,  +  K8.?27)  =  0,  (5) 

lf  =  .0708«.  (6) 

From  (6)  we  obtain 

a  =11.77  rad./sec.2. 

Suppose  the  body  to  begin  to  rotate  from  rest  and  at 
the  time  under  consideration  it  has  been  rotating  2  seconds. 

Then  co  =  2  a  =  23.54  rad./sec. 

Solving  the  remaining  equations,  we  obtain, 

Px  =  _  47.15  Ib.  P'x=  -  27.94  Ib. 

PJ/  =  2.171b.  P;=9.421b. 

Pz  =  8.73  Ib. 

A  negative  sign  indicates  that  the  force  acts  opposite  to 
the  direction  assumed  in  the  figure. 


342 


APPLIED  MECHANICS  FOR  ENGINEERS 


B 


Problem  412.  A  flywheel  3  ft.  in  diameter.  The  cross  section  oi 
the  rim  is  3  in.  x  3  in.  and  is  made  of  cast  iron.  Neglect  the  weight 
of  the  spokes.  This  wheel  is  placed  on  the  axis  of  Fig.  257,  with  its 
center  on  the  axis,  instead  of  the  sphere.  If  the  other  conditions  are 
the  same,  find  the  reactions  of  the  supports. 

Problem  413.  The  sphere  in  Fig.  257  is  replaced  by  a  right 
circular  cast-iron  cone  of  height  one  foot  and  diameter  of  base  one 

foot.  The  vertex  is  placed  at  the  point 
of  attachment  of  the  sphere  (4  in.  from 
the  axis  of  rotation),  and  the  base  is  out- 
ward. If  the  other  conditions  are  the 
same,  find  the  reactions  of  the  supports. 

Problem  414.  In  the  problem  of  the 
sphere  (Art.  181)  find  the  angular  veloc- 
ity at  the  end  of  30  sec.  and  the  reactions 
of  the  supports  for  such  speed. 

Problem  415.  The  two  spheres  of 
Fig.  258  each  weigh  220  Ib.  and  are  12 
in.  in  diameter.  The  rod  connecting 
them  is  inclined  30°  to  the  horizontal  and 
is  rigidly  connected  to  the  vertical  axle 
AB.  Find  the  reactions  at  A  and  B  when  the  system  is  rotating 
uniformly  at  120  r.  p.  in.,  neglecting  the  weight  of  the  rod. 

Problem  416.     The 

spheres  C  and  D  of  Fig. 
259  weigh  respectively  100 
Ib.  and  50  Ib.  The  planes 
containing  their  centers 
and  the  axis  of  the  shaft 
are  90°  apart.  If  they  are 
rotating  at  150  r.  p.  m.  when 
C  is  vertically  above  the 
shaft,  and  no  external 
forces  are  acting  except  the  weights  and  the  reactions  of  the  smooth 


FIG.  258 


FIG.  259 


DYNAMICS  OF  RIGID  BODIES 


343 


bearings,  find  the  reactions  at  A  and  B  when  C  is  (a)  vertically  above 
the  shaft,  (&)  vertically  below  the  shaft. 

Problem  417.  A  uniform  sphere  of  radius  1  ft.  and  weight  2000 
Ib.  is  mounted  on  a  horizontal  axle  of  diameter  4  in.,  round  which 
a  small  cord  is  wound  carrying  a  weight  of  50  Ib.  Neglecting  fric- 
tion, find  the  angular  acceleration  of  the  sphere,  the  linear  accelera- 
tion of  the  weight,  the  tension  in  the  cord,  and  the  velocity  of  the 
weight  at  the  end  of  5  sec. 

SUGGESTION.  Consider  the  motion  of  the  sphere  and  the  weight 
separately,  under  the  action  of  the  forces  acting  on  each. 

Problem  418.  A  rectangular  slab  6  ft.  by  8  in.  by  2  in.  is  sup- 
ported by  an  axis  through  the  slab  perpendicular  to  the  face  6  ft.  by 
8  in.  at  a  distance  of  1  ft.  from  one  end  and  4  in.  from  the  side. 
The  slab  is  pulled  aside  to  a  horizontal  position  and  released.  Find 
the  reaction  on  the  sup- 
port when  the  slab  has 
turned  through  60°  ;  when 
through  90°,  the  weight  of 
the  slab  being  320  Ib. 


Go  =  £0  LB& 


SUGGESTION.  Choose  z- 
axis  as  axis  of  support  and 
ar-axis  along  the  axis  of  the 
slab  at  the  instants  in 
question. 

Problem     419.       Two 

drums  whose  radii  are  r±  = 

16  in.  and  rz  =  12  in.  are 

mounted  as  shown  in  Fig. 

260.     Their    combined 

weight  is  200  Ib.  and  k  = 

14".     The  forces  GI  and  G2  act  upon  the  drum,  as  well  as  journal 

friction  amounting  to  16  Ib.     The  radius  of  the  shaft  is  1  in.     Find 

the  velocities  of  Gl  and  G2  and  the  drums  when  the  point  of  attach- 


FIG.  260 


344 


APPLIED  MECHANICS  FOR  ENGINEERS 


ment  of  the  cord  on  the  small  drum  has  traveled  from  rest  at  A  to  A9 
through  90°.     Neglect  the  friction  at  B. 

182.  The  Compound  Pendulum.  — A  body  rotating  under 
the  action  of  gravity  about  a 
horizontal  axis  is  called  a  com- 
pound pendulum. 

Let  Fig.  261  represent  a  com- 
pound pendulum  rotating  about 
an  axis  through  0  perpendicular 
to  the  plane  of  the  paper.  Let 
the  distance  from  the  axis  of  ro- 
tation to  the  center  of  gravity  of 
the  body  be  d. 

Applying  D'Alembert's  princi- 
FIG.  261  ple>  and  taking  moments  about  0, 


we  have 
or,  since 


—  G-d  sin  0  =  a!Q, 


It  was  shown  in  Art.  120  that  the  tangential  acceleration 
of  a  simple  pendulum  of  length  I  is 

at  =  —  9  sin  0» 
and  hence  its  angular  acceleration  is  ^,  or 

a  =  —  @-  sin  6. 
l 

Hence  the  angular  motion  of   a  compound  pendulum  is 
exactly  the   same    as  that  of  a  simple  pendulum  whose 


DYNAMICS   OF  RIGID  BODIES  345 

length,  I,  is  such  that 

^=i 
*»      I9 

l-*% 
or  ~d 

This  value,  ^1,  is  called  the  length  of  the  equivalent  simple 
pendulum. 

The  time  of  a  small  vibration  of  a  compound  pendulum 
is  therefore 


The  point,  0,  of  the  pendulum,  about  which  rotation 
takes  place,  is  called  the  center  of  suspension. 

The  point   0'  on  OO  such  that   00'  =^f  is  called  the 

d 

center  of  oscillation.  It  is  that  point  at  which  the  whole 
mass  might  be  concentrated  without  changing  the  time  of 
vibration,  the  center  of  suspension  remaining  the  same. 

Problem  420.  A  board  4  ft.  by  1  ft.  by  1  in.  vibrates  about  an 
axis  perpendicular  to  the  4  ft.  by  1  ft.  face  through  a  point  of  the 
board  18  in.  from  the  center.  Find  the  time  of  vibration  for  small 
oscillations,  and  the  length  of  the  equivalent  simple  pendulum. 

Problem  421.  Show  that  the  time  of  vibration  is  the  same  (for 
equal  angles  of  vibration)  for  all  parallel  axes  of  suspension  that  are 
at  equal  distances  from  the  center  of  gravity  of  the  body. 

Problem  422.  A  cast-iron  sphere  whose  radius  is  6  in.  vibrates 
as  a  pendulum  about  a  tangent  line  as  an  axis.  Find  the  period 
of  vibration  and  the  length  of  a  simple  pendulum  having  the  same 
period.  Locate  the  center  of  oscillation. 


346  APPLIED  MECHANICS  FOR   ENGINEERS 

183.  Centers  of  Oscillation  and  Suspension  Interchangeable.  — 
If  k  is  the  radius  of  gyration  of  the  body  about  the  axis 
through  the  center  of  gravity  parallel  to  the  axis  of  sus- 
pension, then 


Therefore 
or 

In  this  formula  dand  I  —  d  enter  in  exactly  the  same  way. 
It  follows  therefore  that  if  0'  were  taken  as  a  center  of 
suspension,  0  would  be  the  center  of  oscillation.  That 
is,  in  a  compound  pendulum  the  centers  of  oscillation  and 
suspension  are  interchangeable.  This  is  easily  verified  ex- 
perimentally. 

Problem  423.     In  a  plane  through  the  center  of  gravity,  (7,  of  a 

k2 
body  circles  of  radii  r  and  —  are  drawn  with  centers  at  C,  where  k  is 

the  radius  of  gyration  about  a  gravity  axis  perpendicular  to  the 
given  plane.  Show  that  the  time  of  vibration  is  the  same  for  all 
axes  of  suspension  perpendicular  to  the  given  plane  and  passing 
through  a  point  on  the  circumference  of  either  circle. 

184.   Experimental  Determination  of  Moment  of  Inertia.— 

The  computation  of  the  moment  of  inertia  of  many  bodies 
is  a  difficult  matter.  It  is  often  convenient,  therefore,  to 
use  an  experimental  method  in  dealing  with  such  bodies. 
The  compound  pendulum  furnishes  a  means  whereby  such 
determinations  may  be  made.  From  Art.  182,  we  find 
that  the  time  of  vibration  of  a  compound  pendulum  is 


DYNAMICS   OF  EIGID  BODIES  347 

This  may  be  written 


Multiplying  both  sides  by  M,  the  mass  of  the  body,  we 
have 


It  thus  appears  that  if  rf,  the  distance  from  0  to  the 
center  of  gravity,  is  known  (the  center  of  gravity  may  be 
located  by  balancing  over  a  knife  edge)  and  also  the  weight 
6r,  and  the  body  be  allowed  to  swing  as  a  pendulum  about 
0  as  an  axis,  t  may  be  determined,  giving  IQ. 

If  Ig  be  desired,  it  may  be  determined  from  the  formula 
(see  Art.  64), 


Problem  424.  The  steel  connecting  rod  of  an  engine  tapers  regu- 
larly from  the  cross-head  end  to  the  crank-pin  end.  Its  length  is  10  ft., 
its  cross  section  at  the  large  end  5.59"  x  12.58"  and  at  the  cross-head 
end  5.59"  x  8.39".  Neglecting  the  holes  at  the  ends,  the  center  of 
gravity  is  64  in.  from  the  cross-head  end.  Compute  its  moment  of 
inertia  about  the  gravity  axis  perpendicular  to  its  plane  face,  and 
find  the  time  of  vibration  when  vibrating  about  the  cross-head  end. 

The  student  should  take  such  a  connecting  rod  as  the  one  in  the 
preceding  problem,  or  other  body,  and  by  swinging  it  as  a  pendulum 
find  its  period  of  vibration.  Compute  the  moment  of  inertia  about 
the  axis  of  suspension  and  about  the  gravity  axis. 

185.  Determination  of  g.  —  From  the  preceding  article 
we  see  that 


=          = 
9      dt*      MM 


348 


APPLIED  MECHANICS  FOR  ENGINEERS 


This  relation  enables  us  to  determine  #,  as  soon  as  we  know 
I0,  M,  arid  c?,  by  determining  the  time  of  vibration  about  the 


point  0.     It  is  evident  that  --  is   a   constant   for   the 

Ma 

body,  when  the  axis  is  through  (9,  and  that  when  once 
determined  accurately,  the  pendulum  might  be  used  to 
determine  g  for  any  locality. 


This  constant,  -^-,  is  known  as  the  pendulum  constant. 
Md 

Problem  425.  A  round  rod  of  steel  6  ft.  long  is  made  to  swing  as 
a  pendulum  about  an  axis  tangent  to  one  end  and  perpendicular  to  its 
length.  The  rod  is  1  in.  in  diameter.  Determine  the  pendulum 
constant. 

Problem  426.  The  center  of  gravity  of  a  connecting,  rod  5  ft. 
long  is  3  ft.  from  the  cross-head  end.  The  rod  is  vibrated  as  a  pen- 
dulum about  the  cross-head  end.  It  is  found  that  50  vibrations  are 
made  in  a  minute.  Find  the  radius  of  gyration  with  respect  to  the 
cross-head  end. 


186.  The  Torsion  Balance. — A  torsion 
balance  consists  of  a  body  suspended 
by  a  slender  rod  or  wire  attached 
rigidly  to  the  body  and  at  the  point  of 
support,  the  center  of  gravity  of  the 
body  lying  in  the  line  of  the  wire  (Fig. 
262). 

Let  OA  be  the  neutral  position  of  a 
line  in  the  surface  of  the  disk  used  as 
a  torsion  pendulum  (Fig.  262). 

Let  the  disk  be  turned  through  an 
angle  0Q  and  released.  The  wire  is 


FIG.  262 


DYNAMICS  OF  RIGID  BODIES  349 

then  twisted  and  exerts  a  twisting  moment  on  the  body, 
tending  to  restore  it  to  the  neutral  position.  Experiment 
shows  that  this  twisting  moment  exerted  by  the  wire  is 
proportional  to  the  angle  of  twist.  Hence,  since 

>  2  mom  =  Ja, 

we  have  for  the  motion  of  the  disk,  starting  from  the  posi- 


tion where  6  =  00, 


CD 


the  minus  sign  being  used  since  o>  -^   is  the  acceleration 

du 

counter-clockwise  and  CO  represents  the  clockwise  twist- 
ing moment. 

.-.  i<0da>  =  -  cede. 

Integrating, 


When          t  =  0,  0=00,  <B  =  0.      .•.£,= 

.:I*=C(Ol-e*).  (2) 

This  equation  shows  that  the  angular  velocity  of  the 
body  is  the  same  for  negative  values  of  0  as  for  the  nu- 
merically equal  positive  values.  The  motion  is  therefore 
periodic,  the  body  vibrating  through  equal  angles  on  both 
sides  of  the  neutral  position. 

Considering  the  motion  from  9  =  00  to  6—  0,  equation  (2) 
may  be  written 

-I— 

de  a 


350  APPLIED  MECHANICS  FOR  ENGINEERS 

Integrating, 


When  t  =  0,  6  =  00.     .  -.  <7a  =  sin-1 1. 


Choosing  —  as  the  value  of  sin"1  1,  we  have 


sin"*—  SB  -r-  —  \/~  •  t. 

Q 

This  equation  shows  that  as  t  increases  sin"1 —  must  de- 
crease. Hence,  since  we  chose  sin"1—  =  —  when  6  =  00, 

a 
the  angle  —  must  decrease  from  —  and  will  therefore  reach 

zero  when  6  =  0. 

Hence  t  =  ?j- -y—         when  0  =  0. 

Therefore  the  time  for  the  body  to  make  a  complete  swing, 
one  way,  is 

— v?-      ;  .  : 

If  m1  is  the  twisting  moment  exerted  by  the  wire  when 
twisted  through  an  angle  0V  m^  =  C0V  and  the  expression 
for  the  period  becomes 


The  time  of  vibration  is  independent  of  the  initial  angu- 
lar displacement. 


DYNAMICS  OF  RIGID  BODIES 


351 


187.  Determination  of  the  Moment  of  Inertia  of  a  Body  by 
Means  of  the  Torsion  Balance.  —  The  moment  of  inertia  of 
a  body  may  be  found  by  suspending  it  by  a  wire  and  ob- 
serving the  time  of  vibration  when  used  as  a  torsion 

balance.     The  constant  (7=  ^  is  a  constant  of  the  wire  or 

ei 

rod  and  depends  upon  the  material  and  diameter.  Know- 
ing this  constant,  it  would  only  be  necessary  to  determine 
the  period  of  vibration  in  order  to  find  /. 

For  practical  purposes,  however,  it  is  desirable  to  elimi- 
nate from  consideration  the  value  ^.  For  this  purpose 

suppose  the  disk  provided  with  a  sus- 
pended platform  rigidly  attached  as 
shown  in  cross  section  in  Fig.  263.  Let 
t  be  its  time  of  vibration  and  I  its  mo- 
ment of  inertia  about  the  axis  of  sus- 
pension. Now  place  on  the  disk  two 
equal  cylinders  H  in  such  a  way  that 
their  center  of  gravity  is  the  axis  of 
suspension.  Let  ^  be  the  period  of 
vibration  of  the  cylinders  and  support 
and  Jj  their  moment  of  inertia.  Fro.  263 

t 2      I 

Then  —  =— .  The  moment  of  inertia  of  the  two  cyl- 
inders with  respect  to  the  axis  of  rotation  is  known; 


If 


call  it 
so  that 


Then 


J-4. 


352  APPLIED  MECHANICS  FOR  ENGINEERS 

This  gives  the  moment  of  inertia  of  the  torsion  balance, 
which,  of  course,  is  a  constant. 

The  moment  of  inertia  of  any  body  L  may  now  be 
determined  by  placing  the  body  on  the  suspended  plat- 
form with  its  center  of  gravity  in  the  axis  of  rotation  and 
noting  the  time  of  vibration.  Calling  the  time  of  vibra- 
tion of  the  body  L  and  the  balance  tz  and  their  moment 
of  inertia  Zj,  we  have 


Let  the  moment  of  inertia  of  L  itself  be  I±,  so  that 


Then 


This  method  may  be  used  in  finding  the  moment  of 
inertia  of  non-homogeneous  bodies,  provided  the  center 
of  gravity  be  placed  in  the  axis  of  rotation. 

Problem  427.  The  moment  of  inertia  of  a  torsion  balance  is 
6300,  where  the  units  of  mass,  space,  and  time  are  respectively  the 
gram,  centimeter,  and  second,  and  its  time  of  vibration  20  sec.  The 
body  L  consists  of  a  homogeneous  cast-iron  disk  3  in.  in  diameter 
and  1  in.  thick.  Find  the  time  of  vibration  of  the  balance  when  L  is 
in  place.  Compute  the  moment  of  inertia  of  the  disk. 

Problem  428.  The  same  balance  as  that  used  in  the  preceding 
problem  is  loaded  with  a  body  L,  and  the  time  of  vibration  is  found 
to  be  30  sec.  Determine  the  moment  of  inertia  of  L. 

188.  Rotating  Body  under  the  Action  of  no  Forces.  —  If  a 
body  is  rotating  about  a  fixed  axis,  the  axis  of  2,  and  is 
acted  upon  by  no  forces,  the  equations  of  Art.  180  become 


DYNAMICS   OF  RIGID  BODIES  353 


0  =  —  co2xM—  ayM,  (1) 

0  =  0,  (3) 

0  =  rfCyzdM  -  aCxzdM,  (4) 

0  =  _  6,2  f  xzd  M-  a  CyzdM,  (5) 

0=oJ..  "  (6) 
From  these  equations  it  follows  that 

«  =  0,       CxzdM=Q,        CyzdM=Q,       x  =  0,  ^  =  0. 

Hence,  if  a  body  is  rotating  about  a  fixed  axis  and  no 
forces  are  acting  on  the  body, 

(1)  the  angular  velocity  is  constant, 

(2)  the    axis   of  rotation  passes  through  the  center   of 
gravity, 

(3)  if  the  axis  of  rotation  is  chosen  as  the  axis  of  z,  then 


CxzdM=  0,  and   CyzdM=  0. 


(When  r xzdM  =  0  and  (yzdM=Q,  the  z-axis  is  a  prin- 
cipal axis  of  the  body.  It  can  be  shown  that  there  are 
three  lines  at  right  angles  to  each  other  through  any  point 
of  the  body  which  are  principal  axes,  and  that  the  ellipsoid 
of  inertia  of  the  body  for  that  point  has  its  axes  lying  on 
these  three  lines.  Compare  Art.  77.) 

189.  Rotation  of  a  Body  about  a  Fixed  Axis  with  Constant 
Angular  Velocity.  —  If  the  angular  velocity  is  constant, 
«=0,  and  the  equations  of  Art.  180  take  simpler  forms. 

2A 


APPLIED  MECHANICS  FOR  ENGINEERS 

111  addition,  at  a  given  instant,  let  the  o>axis  be  chosen 
through  the  center  of  gravity  of  the  body.  Then  y  =  0, 
and  equations  (1)  to  (6)  of  Art.  180  become 


(1) 

2Pi=0,  (2) 

2Z,  =  0,  (3) 

2  in  om  fa  =  a)2  |  yzdM,  (4) 

2  momt-y  =  —  o>2  J  xzdM,  (5) 

2  momt-z  =  0.  (6) 

The  first  equation  shows  that  unless  x  is  zero  there  is  a 
resultant  of  all  the  ^-components  of  the  impressed  forces 
and  that  this  resultant  is 


Equations  (2)  and  (3)  show  that  the  ^-components  of  the 
impressed  forces  either  have  a  resultant  zero  or  else  form 
a  couple,  and  the  same  of  the  2-components. 
A  particular  case  is  that  where 

CxzdM=0,  and   A/zdJf  ==  0, 

as,  for  example,  where  the  a^-plane  is  a  plane  of  symmetry. 
Suppose  in  this  case  the  only  impressed  forces  are  the 
weights  and  the  reactions  of  the  supports,  as  indicated  in 
Fig.  264. 

Equations  (1)  to  (6)  then  become 


(30 


DYNAMICS   OF  BIG  ID  BODIES 


bP-aP'=Q, 


0=0. 


FIG.  264 


From  (20  and  (40,     Pv  =  0,  P'y  =  0. 
From  (30,  P*  =  W. 

From  (10  and  (50,    Px  =  ~^r  W ^ 

a  +  o          a  +  b 


355 

(40 
(50 
(60 


x 

a+b  a+b 

If  the  weight  W  were  counterbalanced  by  an  equal 
upward  force  acting  through  (7,  the  only  forces  acting  on 
the  axis  would  be 


a     - 


a  +  b 


356  APPLIED  MECHANICS  FOB  ENGINEERS 

The  resultant  of  these  forces  is  a  single  force, 


and  if  its  distance  above  the  xy-plane  is 

z'Rx  =  aPi-bP, 

ab 


a+b  a+b 

=  0. 

Therefore,  when  the  #?/-plane  is  a  plane  of  symmetry,  or 
when  j  xzdM=  0  and  J  yzdM '  =  0,  the  effect  of  the  rota- 
tion of  the  body  about  the  axis  of  z  is  to  cause  an  outward 
pull  on  the  axis  of  rotation  in  a  line  through  the  center 
of  gravity  perpendicular  to  the  axis  of  rotation  and  of  the 
same  value,  Mxw2,  that  would  be  caused  by  concentrating 
the  whole  mass  at  the  center  of  gravity. 

Problem  429.  A  wheel  of  weight  100  Ib.  is  mounted  on  an  axle 
and  is  off  center  ^  inch,  the  plane  of  the  wheel  being  perpendicular  to 
the  axis.  Find  the  force  tending  to  bend  the  shaft  when  the  wheel  is 
making  200  r.  p  m. 

Problem  430.  A  thin  rod,  2  ft.  long,  weighing  5  Ib.  is  attached  to 
an  axle  at  an  angle  of  60°.  Find  the  outward 
force  in  magnitude  and  position  in  the  plane  of 
the  rod  and  axle  due  to  the  rotation  when  mak- 
ing 150  r.  p.  m.  If  the  rod  is  joined  to  the  axle 
2  ft.  and  1  ft.  from  the  supports  (Fig.  265),  find 
the  reaction  at  the  supports  due  to  the  rotation. 

Problem  431.    A  steel  disk  3  ft.  in  diameter 
and  1  in.  thick  is  not  perpendicular  to  the  axis 


of  rotation,  but  is  out  of  true  by  r&TT  of  its 
FIG.  265  ......  i 

radius.      Find  the  twisting  couple  introduced 

tending  to   make  the   shaft  wobble. 


DYNAMICS   OF  RIGID  BODIES 


357 


Problem  432.  A  grindstone  weighing  200  lb.,  of  radius  2  ft.,  is 
making  100  r.  p.  m.  Find  the  force  with  which  one  half  of  the  stone 
pulls  on  the  other  half. 

Problem  433.  Neglecting  the  weight  and 
the  tension  in  the  spokes  of  a  rotating  fly- 
wheel, prove  that  the  tension  in  the  rim  is 

f.F 

<J 


FIG.  266 


P  = 


where  <D  =  angular  velocity,  y  =  the  heaviness 
of  the  material,  F  =  the  area  of  the  cross 
section  of  the  rim,  and  r  =  mean  radius  of 
the  rim  (Fig.  266). 

Problem  434.  In  the  preceding  problem, 
suppose  r  =  6  ft.,  F=  10  in.  by  4  in.,  and  the 
wheel  is  made  of  cast  iron.  If  the  tensile  strength  of  the  material  is 
25,000  lb.  per  square  inch,  what  speed  would  be  attained  before  the 
wheel  bursts? 

Problem  435.    Show  that  if  P  is  the  pressure  of  the  side  rod  of  a 

locomotive  on  the 
crank  pin,  r  the  ra- 
dius of  the  crank-pin 
circle,  r'  the  radius 
of  the  driver,  and  v 
the  velocity  of  the 
train, 


O 
FIG.  267 

in  the  lowest  position  (Fig.  267). 


2P-G  = 


Grv2 


when  the  side  rod  is 


Problem  436.  Suppose  the  velocity  of  a  locomotive  to  be  90 
mi.  per  hour,  the  radius  of  the  crank-pin  circle  20  in.,  the  radius 
of  the  drive  wheel  40  in.,  and  the  weight  of  the  side  rod  400  lb. 
Find  the  pressure  on  the  crank  pins  due  to  the  rotation  alone. 


358  APPLIED  MECHANICS  FOR  ENGINEERS 

190.  Rotation  of  a  Locomotive  Drive  Wheel.  —  The  drive 
wheel  of  a  locomotive  (Fig.  268)  may  be  considered  for 
the  present  as  rotating  about  a  fixed  axis.  We  shall  con- 
sider the  effect  of  the  weight  of  the  counterbalance  on  the 
tire  due  to  rotation  only,  on  the 
assumption  that  the  tire  carries  all 
the  weight  of  the  counterbalance. 

NOTE.  It  is  to  be  understood  that  the 
wheel  center  carries  part  of  the  weight  of 
the  counterbalance,  but  a  complete  solu- 
tion of  the  problem  of  the  drive  wheel  is 
beyond  the  scope  of  this  book.  The  above 
assumption  is  therefore  made. 

Let  M  be  the  mass  of  ^  of  tire 
and  p  the  distance  of  its  center  of 
gravity  from  the  center  of  wheel. 

Let  M^  be  the  mass  of  the  counterbalance,  and  /^  the  dis- 
tance of  its  center  of  gravity  from  the  center  of  wheel. 

Then  2  P  =  ft)2  (Mp  +  JKfo) . 

In  particular,  suppose  the  diameter  of  the  tread  of  the  tire 
to  be  80  in.,  distance  of  the  center  of  gravity  of  J  of  tire 
from  center  27  in.,  and  mass  of  J  of  tire  21.  The  mass 
of  the  counterbalance  is  20,  and  the  distance  of  its  center 
of  gravity  from  the  center  of  the  wheel  29  in.  Substitut- 
ing these  values,  we  get  > 

2  P  =  ft)2  [21  (f  J)  +  20  (f|)]  =  95.6  ft)2. 

If  now  we  know  the  speed  of  rotation  of  the  wheel  so 
that  ft)  is  known,  we  may  determine  P.  Let  us  take  ft) 
corresponding  to  a  speed  of  train  of  60  mi.  per  hour. 


DYNAMICS  OF  RIGID  BODIES 


359 


This  gives  o>  =  26.4  radians  per  second  and 
P  =  33,300  Ib. 

Problem  437.  If  the  area  of  the  cross  section  of  the  tire  is  20 
sq.  in.,  find  the  stress  on  the  metal  due  to  rotation  about  the  axis 
under  the  above  assumption. 

If  the  allowable  stress  on  the  metal  is  20.000  Ib.  per  square  inch, 
find  the  speed  of  train  to  cause  this  stress. 

191.  Standing  and  Running  Balance  of  a  Shaft.  —  Let 
weights,  Wv  W2,  TT3,  TF4,  etc.,  be  attached  to  a  shaft,  the 
distances  of  their  centers  from  the  center  of  the  shaft 
being  rv  r2,  r3,  r4,  etc.  (Fig.  269),  it  being  assumed  that 

w* 


FIG.  269 

each  weight  is  symmetric  with  respect  to  a  plane  through 
its  center  and  perpendicular  to  the  axis  of  the  shaft,  so 
that  the  pull  of  the  weight  on  the  shaft  due  to  rotation  is 
through  the  center  of  gravity  of  the  weight  (Art.  189). 

If  the  shaft  is  balanced  when  at  rest  in  any  horizontal 
position,  the  sum  of  the  moments  of  the  weights  about 
the  axis  must  be  zero  for  any  position.  The  moment  of 


360  APPLIED  MECHANICS  FOB  ENGINEERS 

W1  is  W±  times  the  horizontal  component  of  rv  or  it  is 
the  horizontal  component  of  Wlrl  laid  off  along  rr  The 
weights  will  then  be  in  standing  balance  when,  and  only 
when,  the  sum  of  the  horizontal  components  of  the  vec- 
tors ^Wv  r^Wy  r3W3,  etc.,  is  zero  for  any  horizontal  posi- 
tion of  the  shaft.  Hence  the  vector  polygon 

r1F1  +  raIF2  +  r8T78+... 
must  close. 

For  running  balance  the  reactions  at  each  support,  due 
to  the  centrifugal  pull  of  the  weights,  must  annul  or  the 
shaft  will  tend  to  wobble.  These  forces  due  to  rotation 
are  along  the  radii  (Art.  189),  and  if  xr  o?2,  #3,  ...  are  the 
distances  along  the  shaft  of  the  weights  from  one  bearing, 

the  reactions  at  the  other  bearing  are  -*•  r^aPM^  -*  r2<a>2M2, 

L  I 

...,  where  I  is  the  distance  between  the  supports,  and  Mv 
Mv  My  etc.,  are  the  masses  of  the  weights. 

These  reactions  are  therefore  proportional  to  xlr1Wv 
ic2r2TF2,  etc.,  and  are  in  the  directions  of  r^  r2,  etc.,  re- 
spectively. In  order  to  balance,  the  polygon  of  vectors 


must  therefore  close. 

In  order  that  the  reactions  due  to  rotation  balance  at 
the  other  bearing,  the  vector  polygon 

(I  -  *!>!  ffi  +  (*  -  *2>2  Wz  +  -. 

must  close.     But  since  the  vectors  may  be  taken  in  any 
order,  this  polygon  may  be  plotted  in  the  order 


Wl 


DYNAMICS   OF  RIGID  BODIES  361 

which  clearly  will  close  when  the  two  polygons  already 
considered  close. 

There  will  then  be  standing  and  running  balance  when 
two  closed  polygons  can  be  formed  of  vectors  ^  Wv  r2  W^ 
r^Wy  etc.,  and  x^Wv  x2r2W2,  x8rBWB,  etc.,  respectively, 
the  vectors  having  the  directions  of  the  perpendiculars 
from  the  center  line  of  the  axle  to  the  centers  of  gravity 
of  the  corresponding  weights. 

Problem  438.  In  Fig.  269  assume  W^  =  6,  W2  =  5,  Ws  =  4 ; 
TJ  =  4.3,  r2  =  3,  rs  =  6,  r4  =  2 ;  angle  between  rt  and  rs  =  135°,  angle 
between  r3  and  rz  =  90° ;  x1  =  1.5,  xz  =  1.  Find  xs  and  the  value  and 
position  of  W±  for  running  balance. 

Problem  439.  Show  that  two  weights,  given  in  position  and 
magnitude  in  the  same  plane  perpendicular  to  the  axis  of  rotation, 
can  be  balanced  for  running  by  a  third  given  weight  in  that  plane. 
Show  how  to  determine  the  position  of  this  third  weight,  and  show 
that  the  condition  for  standing  balance  is  the  same  as  for  running 
balance  in  the  case  of  weights  all  in  the  same  plane  perpendicular 
to  the  axis  of  rotation. 

Problem  440.  If  two  weights  are  attached  to  the  axis  in  the  same 
plane  containing  the  axis,  show  how  to  find  the  position  of  a  third 
given  weight  that  will  balance  the  two  given  weights  for  running  and 
standing  balance.  If  balanced  for  standing,  will  they  be  balanced  for 
running  ? 

Problem  441.  Show  that  three  weights  cannot  be  balanced  for 
running  unless  they  either  lie  all  in  one  plane  perpendicular  to  the 
axis  of  rotation  or  else  all  in  a  plane  containing  the  axis. 

Problem  442.  Show  that  four  weights  and  the  arms  of  three  of 
them  being  given  and  two  of  the  weights  definitely  located  on  the  axis, 
the  other  weights  may  be  placed  so  as  to  form  standing  and  running 
balance,  and  show  how  the  location  of  the  two  weights  would  be  de- 
termined. Could  this  be  done  in  more  than  one  way? 


362 


APPLIED  MECHANICS  FOR  ENGINEERS 


192.  Rotation  of  Flywheel  of  Steam  Engine.  —  In  Fig.  270 
let  a  belt  run  horizontally  over  a  flywheel,  the  tensions 
being  Pl  and  P2  where  P2  >  Pr  The  effective  steam 
pressure  is  P,  N'  the  pressure  of  the  guides  on  the  cross- 
)iead.  It  is  normal  if  friction  is  neglected. 


FIG.  270 


The  pressure   on   the   crank  pin  is  resolved  into  tan- 
gential and  radial  components,  T  and  N±. 
From  the  relation, 


2  moms  =  «/, 
we  have  for  the  motion  of  the  wheel 


(1) 


It  will  be  assumed  that  the  resistance  of  the  machinery, 
as  shown  by  P2  —  Pv  is  constant. 

When  T=  (P2  —  Pj)  -  ,  a  —  0,  and  the  angular  velocity, 

eo,  is  either  a  maximum  or  a  minimum. 

When  0=0  (Fig.  270),  T=  0,  a  is  negative,  and  <o  is 


DYNAMICS   OF  RIGID  BODIES  363 

therefore  decreasing.     Hence  co  has  a  minimum  value  in 
the  first  quadrant  at  Bl  where  T=  (P2  —  P^)-- 

From  .#!  the  angular  velocity  increases  as  long  as  T 
remains  greater  than  (P2  —  Pj)-.  In  the  second  quad- 
rant as  T  decreases  toward  zero  at  the  dead  point  A,  there 
is  a  point  A1  where  T=  (P2  ~~  P\)~  a^  which  therefore  o> 

has  a  maximum  value.     In  the  same  way  the  points  A1 
and  E!  in  the  third  and  fourth  quadrants  correspond  to 
minimum  and  maximum  values  of  o>. 
Equation  (1)  may  be  written 

ladco  =  Tadd  -  (P2  -  PJrdO.  (!') 

If  «  and  sr  are  the  arcs  passed  over  in  the  crank-pin  circle 
and  the  flywheel  circle  respectively,  then 

adO  =  ds  and  rdd  =  dsr, 

and  equatidn  (!')  becomes  on  integrating  from  an  initial 
value  o)0  to  any  value  o>, 


Since  the  work  done  on  one  end  of  the  connecting  rod 
equals  the  work  done  by  the  other,  neglecting  its  own 
change  in  kinetic  energy, 


CSTds=  £*Pdx, 

*/S0  */^0 

and  hence 

-  a,?)  =  C'Pdx  -  (P2  -  Pjxy  -  <). 

c/X0 


364  APPLIED  MECHANICS  FOE  ENGINEERS 

The  approximate  value  of   I    Pdx  may  be  found  by  read- 

*/#0 

ing  from  the  indicator  card  the  values  of  P  for  successive 
values  of  x  between  the  limits  x  and  #0. 

A  different  treatment  of  the  above  equation  may  be 
obtained  by  considering  that  the  pressure  of  steam  in 
the  cylinder  is  constant  and  equal  to  P1  up  to  the  point 
of  cut-off  and  that  beyond  this  point  the  pressure  varies 
inversely  as  the  volume.  If  we  assume  P  constant  and 
equal  to  P'  to  the  cut-off,  then  the  limits  of  integration 
will  be  regarded  accordingly,  and  we  may  write 


where  xl  is  the  value  of  x  at  the  cut-off. 

Beyond  the  cut-off  P  varies  inversely  as  the  volume  of 
steam  in  the  cylinder,  or 


x 
Then  from  the  point  of  cut-off  to  any  value  of  x, 


=  XlP<  lQg.(j-)-  (P,  -  AX*'  - 


If  P[  is  the  mean  effective  pressure,  we  have,  considering 
the  work  done  on  the  flywheel  for  the  motion  from  B  to 
A,  neglecting  friction  and  the  change  in  kinetic  energy  of 
the  connecting  rod, 

1  J(«J  -  *  D  =  2  a  A  -  (P,  -  PJwr. 

Problem  443.  In  Fig.  270,  given  a  =  18  in.,  I  =  6  ft.,  r  =  3  ft., 
/,  =  2000,  o)B  =  4:  IT  rad./sec.  ;  if  the  total  piston  pressure  is  20,000  Ib. 


DYNAMICS   OF  RIGID  BODIES 


365 


up  to  cut-off  at  \  the  stroke,  show  that  the  mean  piston  pressure  is 
13,991  Ib.  What  constant  resistance,  Pa  —  Pv  will  leave  the  angular 
velocity  the  same  at  the  end  of  one  stroke  ? 

Problem  444.  In  Problem  443  if  <t>B  =  6  TT  rad./  sec.,  what  con- 
stant resistance  will  change  co  to  2?r  rad./sec.  in  100  revolutions,  with 
the  given  steam  pressure  ? 

Problem  445.  Show  that  under  the  conditions  given  in  Problem 
443  the  minimum  value  of  co  comes  before  cut-off,  and  find  its  value. 

Problem  446.  Find  the  maximum  value  of  co  in  the  second  quad- 
rant with  the  conditions  of  Problem  443. 

Problem  447.     In  Problem  443  find  the  value  of  co  when  &  =  90°. 

193.  Rotation  and  Translation,  —  Let  0  be  a  point  in  a 
body  having  plane  motion,  i.e.  translation  and  rotation  in 
a  plane,  and  let  the  angular 
velocity  and  angular  ac- 
celeration of  the  body  be  t» 
and  «  respectively  (Fig. 
271). 

At  any  instant  choose 
axes  with  origin  at  0  and 
the  #-axis  along  the  vector 
of  a,  the  linear  acceleration 
of  the  point  0.  Any  element  of  the  body,  distant  r  from 
0,  then  has  the  acceleration  a  parallel  to  the  #-axis,  a 
tangential  acceleration  ra,  and  a  radial  acceleration  ro)2, 
and  the  corresponding  effective  forces  acting  on  dM  are 
adM,  radM,  and  rMM. 

Applying  D'Alembert's  principle,  taking  the  2-axis 
through  0  perpendicular  to  the  plane  of  motion, 

2J^  =   \(adM—  rco2dMcos  </>  —  radM  sin  <£), 


FIG.  271 


366 


APPLIED  MECHANICS  FOR  ENGINEERS 


mom<2 
which  reduce  to 


<  =  C(ra.dM  cos  <f>  - 
=  J 


—  yadM), 


2  momiz  —a.1.—  dyM. 
Study  these  equations  when  x  =  y  =  0. 


(2) 
(3) 


194.  The  Connecting  Rod.  —  The  connecting  rod  of  an 
engine  has  a  circular  motion  at  one  end  while  the  other 
end  moves  backward  and  forward  in  a  straight  line.  We 
shall  consider  the  motion  relative  to  the  engine  and  shall 
assume  that  the  flywheel  is  of  sufficient  weight  to  give 
the  crank  a  motion  sensibly  uniform.  It  will  be  con- 
venient to  regard  the  motion  of  the  connecting  rod  as  con- 
sisting of  a  rotation  about  the  crosshead  end  while  that  end 
is  moving  in  a  straight  line. 

In  Fig.  272  let  A  be  the  crosshead  and  0  the  center  of 
the  crank-pin  circle.  Let  I  be  the  length  of  the  connect- 
ing rod,  and  r  the  radius  of  the  crank-pin  circle.  If  we 
neglect  friction,  the  only  forces  acting  on  the  connecting 


FIG.  272 


DYNAMICS   OF  RIGID  BODIES  367 

rod  at  A  are  N',  the  pressure  of  the  guides,  and  P  ',  the 
pressure  exerted  by  the  piston  rod.  The  force  exerted  on 
the  connecting  rod  by  the  crank  pin  has  been  resolved 
into  its  normal  and  tangential  components  N±  and  T,  re- 
spectively. Suppose  coj  to  be  the  constant  angular  velocity 
of  the  crank,  and  suppose  the  angular  velocity  of  the  rod 
to  be  represented  by  o>  and  the  angular  acceleration  by  a. 
Let  a  be  the  linear  acceleration  of  the  crosshead. 

The  equations  of  the  preceding  article  applied  to  the 
connecting  rod  become 

P'  -  Tsin  6-^^0*6  =  Ma  -<£xM-ayM,    (1) 
-  N'  -  &  +  JVi  sin  6  -  T  cos  0  =  -<iPyM+  otiM,  (2) 

NJ  sin  (0  +  <£)  -  Tl  cos  (#  +  <£)-  1  &l  cos  </> 

=  ol,  -  ayM.  (3) 

Before  these  equations  can  be  solved  for  the  reactions,  the 
values  of  &>,  a,  and  a  must  be  known.  These  values  can 
be  expressed  in  terms  of  o>r 

From  the  figure, 

r  sin  6  =  I  sin  0. 

Differentiating, 


But  =  ® 

a£ 

.  *.  r^  cos  0  =  l(o  cos  <£, 


or 


Differentiating  again,  we  obtain 


368  APPLIED  MECHANICS  FOB   ENGINEERS 

To  obtain  a  notice  that  any  two  points  of  the  rod  have 
velocities  whose  components  along  the  rod  are  equal. 
Hence  if  vl  is  the  velocity  of  the  crank  pin  and  v  is  the 
velocity  of  the  crosshead, 

v  cos  <f>  =  vl  sin  (0  +  <£)•  (Fig.  272) 

.  •.  v  =  v1  (sin  0  +  cos  0  tan  <£). 


dt  dt  dt  dt 

cos0 


r 

v1\- 

[_ 


COS  (£  COS2</> 

or  a  =  _[WlCOS(e  +  <|,)co8«|>  +  «cos0].  (6) 


The  value  of  o>  from  equation  (4)  may  be  substituted  in 
equation  (6)  and  thus  the  value  of  a  determined. 

Problem  448.  The  connecting  rod  given  in  Problem  424,  Art. 
184,  is  in  use  on  an  engine  whose  crank  has  a  constant  angular 
velocity  of  26  radians  per  second.  The  length  of  the  crank  is  2  ft., 
the  effective  steam  pressure  on  the  piston  is  16,000  Ib.  Use  the  values 
of  M,  /,  x,  and  y  from  Problem  424.  Find  the  values  of  N1,  Nv  and  T 
when  6  =  30°. 

Problem  449.  Show  that  w  has  its  greatest  numerical  value  when 
0  =  0  and  0  =  TT,  and  its  least  numerical  value  when  0  =  —  and  0  =  —  . 
What  are  these  greatest  and  least  values? 

Problem  450.  Find  what  values  of  0  will  make  a  a  maximum  or 
minimum.  Locate  the  crosshead  for  these  values. 

Problem  451.  Find  values  for  T,  N',  Nv  and  the  resultant  pressure 
on  the  crank  pin  when  0  =  TT  and  when  0  =  0.  Use  the  above  data. 

Problem  452.  Assume  a  force  of  friction  F  acting  on  the  cross- 
head,  such  that  F  =  .06  N'.  In  the  above  case  when  0  =  30°,  what  is 
the  value  of  F,  N',  Nv  and  T? 


DYNAMICS   OF  RIGID   BODIES 


369 


FIG.  273 


Problem  453.     Suppose  the  steam  pressure  zero,  find  T,  N',  Nv 
and  the  resultant  crank-pin  pressure,  if  o^  is  the  same. 

195.  Angular  Momentum  or  Moment  of  Momentum.  —  If 

the  velocity  of  a  particle  is  resolved  into  two  components, 
one  of  which  is  parallel  to  a  given 
line  and  the  other  in  a  plane  perpen- 
dicular to  the  line,  the  product  of  the 
latter  component,  the  perpendicular 
distance  of  this  component  from  the 
given  line,  and  the  mass  of  the  particle 
is  called  the  moment  of  momentum  or 
angular  momentum  of  the  particle  with 
respect  to  the  given  line.  Thus,  if 
vl  is  the  component  of  the  velocity  in 
the  plane  perpendicular  to  the  given  line,  d  the  distance 
of  this  component  from  the  given  line,  and  m  the  mass  of 
the  particle,  then  its 

angular  momentum  =  mv^d.  (Fig.  273) 
If  the  given  line  be  taken  as  the  axis  of  #,  the  component 
v1  may  be  resolved  into  components  vy  and  vz  parallel  to 
the  y-  and  2-axes,  respectively.  Then,  since  the  moment 
of  any  vector  is  equal  to  the  sum  of  the  moments  of  its 
components,  we  have 

angular  momentum  of  the  particle  with  respect  to  #-axis 
=  (yvz-zvy)m. 

If  dm  is  the  element  of  mass  of  any  body,  then  the  angu- 
lar momentum  of  the  body  about  any  line  is  defined  to  be 
the  sum  of  the  angular  momenta  of  its  particles  about  that 
line.     Thus,  representing  the  angular  momenta  of  a  body 
2s 


370  APPLIED  MECHANICS  FOR  ENGINEERS 

about  the  #-,  #-,  and  z-axes  by  Ax,  hy,  hz,  respectively,  we 
have 


=\ 


=  ( 


196.  Torque  and  Angular  Momentum.  —  Let  dm  be  the 
mass  of  an  element  of  a  body,  and  a^  ay,  az,  respectively, 
the  components  of  its  acceleration  parallel  to  the  axes. 

Using  D'Alembert's  principle,  we  may  equate  the  sum 
of  the  moments  of  the  effective  forces  to  the  sum  of  the 
moments  of  the  impressed  forces. 

The  effective  forces  acting  on  dm  are  axdm,  avdm,  azdm, 
respectively  parallel  to  the  a?-,  y-,  and  2-axes.  Hence  call- 
ing the  torques  due  to  the  impressed  forces  about  the  axes 
respectively  Tx,  Ty,  Tz,  we  have 

Tx  =(yae  -  za,y)dm, 


(  (zax  —  xaz 


AT                            dx  dy  dz 

Now               V*  =  TS  vy=~T^  Vz  =  ^~S 

dt  dt  dt 

dv  dv  dVz 


Also  —  (yv,  -  zvy}  =  vvvz  +  yaz—  vzvy  -  zay  =  yaz  -  zav. 

dt 


Since  the  derivative  of  a  sum  of  terms  is  the  sum  of  the 
derivatives  of  the  separate  terms,  and  the  integral  is  the 


DYNAMICS  OF  RIGID  BODIES  371 

limit  of  a  summation,  we  may  write  this  latter  equation 


Likewise  Ty  , 

at 

and  r,  =      '- 


That  is  :  2%e  moment  of  impressed  forces  about  any  axis  is 
equal  to  the  rate  of  change  of  the  angular  momentum  about 
that  axis. 

A  corollary  of  this  theorem  is  that  when  the  moment  of 
impressed  forces  acting  on  a  body  about  an  axis  is  con- 
stantly equal  to  zero,  the  angular  momentum  about  that 
axis  is  constant. 

197.  Moment  of  Momentum  of  a  Body  with  One  Point 
Fixed  in  Terms  of  Angular  Velocity.  —  If  a  body  in  motion 
has  one  point  fixed,  its  motion  at  any  instant  is  one  of 
instantaneous  rotation  about  an  axis  passing  through  the 
fixed  point.  Take  the  fixed  point,  0,  of  the  body  as 
origin  of  coordinates.  The  angular  velocity  about  the 
instantaneous  axis  can  be  resolved  into  component 
angular  velocities,  o>x,  &>„,  G>Z,  about  the  axes  (Art.  132). 

The  velocity  vx  or  —  is  then  due  to  rotation  about   OY 
at 

and  OZ.  Figure  274  shows  the  component  parallel  to  OX 
of  the  velocity  due  to  rotation  about  OZ  to  be 

—  rld)z  cos  /3  or  —  ya>z. 


372 


APPLIED  MECHANICS  FOB  ENGINEERS 


In  the  same  way  the  z-component  of  the  velocity  due  to 
rotation  about  OYis  za),,. 


/u 


FIG.  274 


Similarly 
and 


dx 
dt 

dy_ 

dt 


z  —  za> 


Substituting  these   values  in  the  expressions  for  Ax,  we 
obtain 

^x  =  J  [j/G/^x  -  xo>v)  ~  z(xa)*  ~  2ft)x)]c?w 

—  T(^2  _}_  22)ft)xc?m  —  J  xyuydm  —  J  xzcozdm, 
or  Tix  =  Ix<ox  —  a>yj  xydm  —  cozj  xzdm. 

Similarly,   hy  =  Iya>y  —  coz(  yzdm  —  coxl  yxdm, 
and  hz  =  Iz(oz  —  (ox\  zxdm  —  coyj  zydm. 


DYNAMICS   OF  RIGID  BODIES  373 

The  only  body  whose  motion  we  shall  study  in  this  book 
is  a  uniform  body  in  the  form  of  a  solid  of  revolution.  If 
one  of  the  axes  be  taken  as  the  axis  of  the  body,  then  all 
of  the  products  of  inertia  vanish  and  the  expressions  for 
the  angular  momenta  become 

kt      =       IXWX,  ky       =       Iy(Dy,  k,,       =       IjOg. 

198.  Vector  Representation  of  Angular  Momentum.  —  Sim- 
ilar to  the  representation  of  angular  velocity  by  a  vector, 
angular  momentum  about  any  line  may  be  represented  by 
a  vector  along  the  line.     Since  in  the  special  case  we  shall 
study  the  angular  momentum  about  any  axis  used  is  pro- 
portional to  the  angular  velocity  about  that  axis,  it  is  clear 
that  the  angular  momentum  may  properly  be  treated  as 
a  vector  in  the  way  defined.       The  signs  of  the  angular 
momenta  will  be  the  same  as  those  of  the  corresponding 
angular  velocities,  and  the  vectors  will  be  laid  off  in  the 
same  way  as  for  angular  velocities. 

199.  Kinetic  Energy  of  a  Body  with  One  Fixed  Point.  — 
The  kinetic  energy  of  a  body  at  any  instant  is  the  sum  of 
the  kinetic  energy  of  its  particles,  or 

K.  E. 

In  Art.  197  it  was  proved  that 

vx  =  zcoy  —  ywz,     vy  =  xcoz  —  z<0x,     vz  =  y&x  —  xa>v. 

The  substitution  of  these  values  in  the  expression  for 
kinetic  energy  gives 

K.  E. 


374  APPLIED  MECHANICS  FOR  ENGINEERS 

-      \zya>ya>z 


or,  "^iVi+-l  +  i1."! 

—  uyuz\zydm  —  Wa-w-j,  \  xzdm  -  <A 

For  the  special  case  of  the  body  of  revolution  with  one 
of  the  coordinate  axes  coinciding  with  the  axis  of  the 
body,  or  for  any  body  when  referred  to  principal  axes, 


\l 

\     1, 


200.  Vector  Rate  of  Change  Due  to  Rotation.  —  Let  h  be 
a  vector  with  point  of  application  at    0.     Suppose  the 

vector  is  changing  with  the  time 
in  magnitude  and  direction  but 
keeps  the  point  of  application  the 
same.  Let  the  vector  change  in 

length  from  h  to  h  +  AA  in  time  A£  and  in  that  time  turn 
through  the  angle  A0.  The  rate  of  change  in  the  original 
direction  of  the  vector  is  then 

Limit  (A  +  A^)cos  A0  -  h 


and  the  rate  of  change  perpendicular  to  the  original  direc- 
tion of  the  vector  is 


The  first  of  these  quantities  may  be  written 
Limit!"-  *fl  -OMAg>+^  cog 


DYNAMICS  OF  RIGID  BODIES 


375 


Lim 


Lim 


,     S1DT   .    A0     A0 

-A'-ArsmT'^ 


-ft.  1,0=1 


i  dh 


dt      dt 


rfA 

dc 


The  second  is 


T  •     /T,  •  ATA 

Lim  (A  +  AA)  — 

Ac/     Ac 


where  CD  is  the  angular  velocity  of  rotation  of  the  vector. 
Hence  the  rate  of  change  in  the  direction  of  the  vector  is 

dh 


and  the  rate  of  change  at  right  angles  to  the  vector  is 

hco. 

201.  Kate  of  Change  of  Angular  Momentum  of  a  Body  Due 
to  Rotating  Axes.  —  Let  the  components  of  the  angular  mo- 
mentum of  a  body  with  fixed  point  0  with  respect  to  axes 
OJf,  OY,  OZ  be  respectively  hv  A2,  A3,  and  let  this  frame 
of  axes  coincide  at  a  given  instant  with  a  fixed  frame  OXV 


0 


376  APPLIED  MECHANICS  FOR  ENGINEERS 

OYV  OZV  but  be  rotating  about  the  fixed  frame  with 
angular  velocities  o)v  o>2,  o>3. 

Lay  off  the  vectors  hv  A2,  A3  on  the  moving  axes  (Fig. 
276).     Since  the  moving  axes  at  the  given  instant  coincide 

with  the  fixed  axes,  the  com- 
ponents of  the  momentum  are 
the  same  for  the  fixed  and 
moving  axes  at  that  instant. 
The  rate  of  change  of  angular 
momentum  along  the  fixed 

^ f  1  x    axes   is   then    given   by   the 

y  method     of     the     preceding 

^  article.     Thus  along  OX1  the 

rate  of  change  is  — -1,  due  to 

j.  at 

FIG.  276  ,         ,  •     z         rnu*     •     • 

the  change  in  hr  This  is  in- 
creased by  A3o>2  by  the  rotation  of  h3  about  the  ?/-axis  and 
by  —  A2o>3  by  the  rotation  of  A2  about  the  z-axis.  Thus  the 
rate  of  change  of  angular  momentum  about  the  fixed  x- 
axis  is 

dh1_h(0  +ho) 
dt 

Similarly  about  the  fixed  ?/-axis  the  rate  of  change  is 

—2  —   k^  +    AjWg, 

at 
and  about  the  fixed  s-axis  the  rate  of  change  is 

•  ^i«2  +  V>r 


DYNAMICS   OF  RIGID  BODIES 


377 


Since  the  moment  of  impressed  forces  about  any  axis  is 
equal  to  the  rate  of  change  of  the  angular  momentum  about 
that  axis,  we  may  write 


az 


tdm. 


/ 

ay 


T. 


202.  Motion  of  the  Center  of  Gravity  of  Any  Body.  —  Let 

dm  be  the  mass  of  any  element  of  a  body  and  (z,  y,  z)  the 
coordinates  of  dm  (Fig.  277).     The 
effective  forces  acting  on  dm  are  re- 
spectively 

axdm,  aydm,  and  azdm. 

Equating  the  components  of  the  im- 
pressed and  effective  forces  acting  on 
the  body,  parallel  to  the  z-axis,  we 
have 

Now   if   x  is   the  abscissa   of   the 
center  of  gravity  of  the  body, 

Mx  = 
Differentiating, 


FIG.  277 


and 


378 


APPLIED  MECHANICS  FOE  ENGINEERS 


;  =  Ma 


Therefore,  if  ax  is  the  ^-component  of  the  acceleration  of 
the  center  of  gravity  of  the  body,  equation  (1)  may  be 
written 


In  like  manner, 


Hence,  for  any  motion  of  a  body  the  acceleration  of  the 
center  of  gravity  is  the  same  as  if  the  whole  mass  were  con- 
centrated there  and  all  the  impressed  forces  were  applied 
there  parallel  to  their  original  directions.  (See  Art.  193.) 

203.  Gyroscope.  Simplest  Case  —  Consider  a  body  in  the 
form  of  a  solid  of  revolution  turning  with  uniform  angular 
velocity,  &>,  about  its  geometric  axis,  which  is  horizontal, 

while  at  the  same  time 
this  axis,  passing  always 
through  a  fixed  point, 
turns  with  uniform  an- 
gular velocity,  H,  about 
the  vertical  (Fig.  278). 
The  problem  is  to  de- 
termine the  forces  neces- 
sary to  maintain  this 
motion. 

Let  00  be  the  axis  of  the  body,  moving  with  the  body. 
Choose   fixed   axes    OX,    OY,    OZ,  so  that  the  z-axis 
coincides  with  0  O  at  the  given  instant. 

Let  OA  be  the  horizontal  axis  perpendicular  to  00. 
At  the  given  instant  OA  coincides  with  OY. 


FIG.  278 


DYNAMICS   OF  RIGID  BODIES  379 

Denote  the  moment  of  inertia  of  the  body  about  OO  by  C, 
and  that  about  OA  or  OZ  by  A.  Then  from  Art.  197  the 
angular  momenta  about  00,  OA,  and  OZ  are  respectively 

Ax  ==  Coo,     A2  =  0,     A3  =  A£l. 

These  values  of  h^  A2,  and  A3,  while  they  are  the  values  of 
the  angular  momenta  about  the  fixed  axes  OX,  OY,  OZ, 
at  the  given  instant,  are  not  values  which  hold  in  general, 
and  cannot  therefore  be  differentiated  to  obtain  the  torques 
about  these  axes.  The  method  of  Art.  201  applies  here 
where 

G)j  =  0,       O>2  ==  0?       ^3  =  Q" 

We  may  write  then 


°-  (3) 

For  the  motion  of  the  center  of  gravity  of  the  body,  since 
the  center  of  gravity  is  moving  uniformly  in  a  horizontal 
circle  and  is  at  the  given  instant  on  the  #-axis,  the  acceler- 
ation is  directed  toward  the  point  of  support  along  the 
a>axis.  Hence 


where  b  is  the  distance  from  the  point  of  support  to  the 

center  of  gravity. 

.-.  2JT,  =  -  jtf"  Ml8,  (4) 

2F,  =  0,  (5) 

2Z<=0.  (Art.  202)     (6) 


380  APPLIED  MECHANICS  FOR  ENGINEERS 

The  impressed  forces  necessary  to  satisfy  equations  (1) 
to  (6)  may,  under  certain  conditions,  be  only  the  weight 
of  the  body  and  the  reaction  of  the  support.  If  these  are 
the  only  impressed  forces,  equation  (2)  becomes 

Wb  = 


where  W  is  the  weight  of  the  rotating  body.  The  re- 
maining equations  are  then  satisfied,  so  that  the  necessary 
and  sufficient  condition  that  the  body  have  the  motion 
described  is 

=  Wb. 


The  motion  of  the  body  around  the  vertical  axis  is 
called  precession.  When  the  angular  velocity  around  the 
vertical  axis  is  constant,  the  precession  is  said  to  be  steady. 

Problem  454.  If  the  rotating  body  is  a  thin  disk  2  ft.  in  diam- 
eter fastened  to  a  rod  of  negligible  weight  which  passes  through  the 
center  of  the  disk  perpendicular  to  its  faces,  the  center  of  the  disk 
being  6  in.  from  the  point  of  support,  what  will  be  the  rate  of  pre- 
cession in  a  horizontal  plane  when  the  angular  velocity  of  rotation 
of  the  disk  about  its  axis  is  300  r.  p.  m.  ?  Ans.  9.79  r.  p.  m. 

Problem  455.  How  would  the  rate  of  precession  vary  with  the 
radius  of  gyration  of  the  disk  if  the  angular  velocity  o>  be  kept  un- 
changed ? 

204.  The  Gyroscopic  Couple.  —  Equation  (2),  Art.  203 
shows  that  when  a  body  of  revolution  is  rotating  uniformly 
about  its  axis  and  this  axis  is  rotating  uniformly  about 
an  axis  at  right  angles  to  it,  there  is  acting  on  the  body 
at  right  angles  to  the  other  two  axes  a  couple  whose  value 
is  the  product  of  the  moment  of  inertia  of  the  body  about 
its  axis  of  revolution  and  the  two  angular  velocities.  This 


DYNAMICS   OF  RIGID  BODIES  381 

couple  is  called  the  gyroscopic  couple.  In  the  case  of  the 
body  discussed  in  Art.  203,  the  gyroscopic  couple  is  Ccofl. 
Since  action  and  reaction  are  equal  and  opposite,  the 
rotating  body  will  resist  the  turning  of  its  plane  of  rota- 
tion with  a  couple  equal  and  opposite  to  the  gyroscopic 
couple.  This  finds  application  in  the  effect  of  rapidly 
rotating  wheels  of  cars,  automobiles,  etc.,  when  turning 
around  curves.  In  Fig.  278  if  the  disk  represents  a  wheel 
of  a  car  rolling  in  the  direction  indicated  by  o>  and  at  the 
same  time  turning  about  OZ,  the  rotation  about  OZ  would 
be  in  the  opposite  direction ;  i.e.  fl  would  be  negative 
in  the  formula  Tv  =  CcoQ.  The  gyroscopic  couple  acting 
on  the  wheel  would  then  be  in  the  opposite  direction,  or 
about  OF  in  the  direction  from  OX  to  OZ.  The  couple 
that  the  wheel  would  exert,  tending  to  upset  the  car,  would 
then  be  about  0  Y  in  the  direction  from  OZ  to  OX. 

Problem  456.  A  ship  carries  a  cast-iron  flywheel  whose  rim  is 
6  ft.  outside  diameter,  4  in.  thick,  and  18  in.  wide.  When  it  is 
making  3  revolutions  per  second,  its  axis  is  turned  about  an  axis 
through  the  plane  of  the  wheel  with  unit  angular  velocity.  Find  the 
moment  of  the  couple  that  tends  to  turn  the  wheel  about  an  axis 
perpendicular  to  these  two  axes. 

Problem  457.  A  solid  cast-iron  disk  3  ft.  in  diameter  and  3  in. 
thick  revolves  about  its  axis,  making  3000  revolutions  per  minute. 
At  the  same  time  it  is  made  to  turn  about  an  axis  in  its  plane  at 
the  rate  of  2  revolutions  per  minute.  Find  the  magnitude  of  the 
couple  tending  to  rotate  the  disk  about  an  axis  perpendicular  to  these 
two  axes. 

Problem  458.  A  locomotive  is  going  at  the  rate  of  40  mi.  per 
hour  around  a  curve  of  600  ft.  radius.  The  diameter  of  the  drivers 
is  80  in.,  and  a  pair  of  drivers  and  axle  have  a  moment  of  inertia 


382  APPLIED  MECHANICS  FOR  ENGINEERS 

about  the  axis  of  rotation  of  1660.  What  is  the  magnitude  of  the 
couple  introduced  by  the  precessional  motion  of  this  pair  of  wheels  ? 
Give  the  direction  in  which  it  acts.  Does  it  tend  to  make  the  loco- 
motive tip  inward  or  outward  ? 

Problem  459.  A  car  pulled  by  the  locomotive  in  the  preceding 
problem  has  four  pairs  of  wheels.  The  moment  of  inertia  of  each 
pair  of  wheels  and  their  connecting  axle,  with  respect  to  the  axis 
of  rotation,  is  38.  What  is  the  magnitude  of  the  precessional  couple 
acting  upon  the  whole  car  ? 

Problem  460.  The  flywheel  of  an  engine  on  board  a  ship  makes 
300  revolutions  per  minute.  The  rim  has  the  following  dimensions : 
outside  radius  4  ft.,  inside  radius  3£  ft.,  width  12  in.  The  ship  rolls 
with  an  angular  velocity  of  £  a  radian  per  second ;  find  the  torque 
acting  on  the  ship  due  to  the  gyrostatic  action  of  the  flywheel. 

205.  Car  on  Single  Rail.  — An  interesting  application 
of  the  gyroscope  has  been  made  recently  in  England.  A 
car  is  run  upon  a  single  rail,  and  is  held  upright  by  means 
of  rapidly  rotating  flywheels.  Each  car  contains  two  of 
these  wheels  rotating  in  opposite  directions,  at  the  rate  of 
8000  revolutions  per  minute. 

Any  tendency  of  the  car  to  tip  over,  either  when  running 
or  standing  at  the  station,  is  righted  by  the  gyroscopic 
action  of  the  flywheels.  The  experimental  car  was  so 
successful  in  operation  that  it  maintained  itself  in  an 
upright  position  even  when  loaded  eccentrically.  The 
action  of  the  flywheels  is  such  as  to  place  the  center  of 
gravity  of  the  car  and  load  directly  over  the  rails.  See 
Engineering ,  June  7,  1907. 

Another  practical  application  of  the  gyroscopic  couple 
is  to  be  found  in  Schlick's  "stabilizer  "  for  steadying  ships. 


DYNAMICS   OF  EIGID  BODIES 


383 


206.  Gyroscope.  Inclined  Axis.  —  Let  the  angular  veloc- 
ity of  the  body  relative  to  its  own  axis,  OC,  be  &>  at  any 
instant,  the  angular  velocity  with  which  the  plane  con- 
taining the  axis  turns  about  the  vertical  be  H,  and  the 
angle  which  the  axis  of  the  body  makes  with  the  vertical 
be  0.  The  component  about  00  of  the  total  angular 
velocity  of  the  body  is  then  o>  +  the  component  of  fl 
about  OC  or  o>  -f  H  cos  6. 

The  angular  velocity  &>  is  called  the  velocity  of  spin. 

Let  —  =  ft>'.     Then  to'  is   the  angular  velocity  of  the 

axis  of  the  body  in  the  vertical  plane  containing  the  axis. 

Choose  fixed  axes  so  that 
at  the  instant  in  question 
the  zz-plane  contains  the 
axis  of  the  body,  the  x- 
axis  being  horizontal  (Fig. 
279).  Let  OA,  OB,  OO  be 
a  set  of  moving  axes,  of 
which  00  is  the  axis  of  the 
rotating  body,  OB  is  hori- 
zontal, and  OA  is  at  right 
angles  to  00  and  OB.  Let 
OAV  OBV  OCl  be  fixed 
axes  which  at  the  given  instant  coincide  with  OA,  OB, 
00  respectively.  The  frame  of  moving  axes  OABO  then 
has  angular  velocities  about  the  fixed  axes  OAV  OBV  OO1 
respectively 

Wj  =  —  fl  sin  6,     ft)2  =  a)',     G>3  =  H  cos  6. 
The  angular  velocity  fl  may  be  resolved  into  components 


384  APPLIED  MECHANICS  FOR  ENGINEERS 

about  OA,  OB,  OO  respectively 

-  fl  sin  0,    0,    H  cos  0. 

The  angular  velocities  of  the  body  about  the  axes   OA, 
OB,  OO  respectively  are  therefore 

—  fl  sin  6,    o)f,   and  o>  -f-  fl  cos  6. 

If  the  moments  of  inertia  of  the  body  about  the  axes 
OA,  OB,  OO  are  respectively  A,  B,  C,  or,  because  of  sym- 
metry, A,  A,  C,  the  angular  momenta  of  the  body  about 
these  axes  are  respectively 

h1  =  -A£lsiii0,   h2  =  Aco',    hs  =  (7(&>  +  O  cos  0). 

The  discussion  will  be  divided  into  two  cases  :  (a)  steady 
precession,  (5)  unsteady  precession. 

(a)  Steady  Precession.  Suppose  the  angle  6  and  the 
angular  velocities  co  and  £1  are  all  constant.  For  this  case 
e»'  =  0  and 

\  =  -  A£l  sin  0,         h2  =  0,         hs  =  (7(ft>  +  fl  cos  6), 
co1  =  —fl  sin  6,  G>2  =  0,         &>3  =  II  cos  0. 

Also,  equating  the  torques  and  the  rates  of  change  of  the 
angular  momenta  about  the  axes  OAV  OBV  OOV  we  have 


0,  (1) 


=  Ofl  sin  0(a>  +  ii  cos  0)  -  ^Ifl2  sin  0  cos  0,         (2) 


DYNAMICS   OF  RIGID  BODIES  385 

It  is  possible  for  this  motion  to  take  place  under  the 
action  of  the  weight  and  the  reaction  of  the  support  as 
the  only  impressed  forces  for  certain  related  values  of 
&>,  H,  and  the  constants  of  the  body.  For  if  only  the 
weight  and  the  support  act  on  the  body,  then,  since  the 
center  of  gravity  moves  as  if  all  the  forces  were  concen- 
trated there,  the  only  additional  conditions  to  be  satisfied 

are 

Xr  =  -Mb  sin  0ft2,  (4) 

rr=0,  (5) 

Zr=W,  (6) 

where  Xr,  Yr,  Zr  are  the  components  of  the  reaction  of  the 
support  along  the  fixed  axes  at  the  given  instant,  and  b  is 
the  distance  from  the  point  of  support  to  the  center  of 
gravity  of  the  body. 

With  only  the  weight  and  the  reactions  of  the  support 
as  impressed  forces,  the  torques  Ta  and  Tc  are  both  zero 
and  Tb  =  Wb  sin  6.  All  of  the  conditions  of  motion  would 
then  be  satisfied  if,  from  equation  (2), 

Wb  sin  6  =  C£l  sin  <9(&>  +  ft  cos  0)  -  AW  sin  0  cos  0, 
or  (A  -  0)  cos  0ft2  -  <7o>ft  +  Wb  =  0. 

The  values  of  II  that  satisfy  this  equation  are 


2ft)2-4  Wb(A-  6Qcos0 
2(JL-  <7)cos0 

There  are  therefore  two  solutions,   one  solution,  or  no 
solution  according  as 


-  <7)  cos  0. 

^x, 

2c 


386  APPLIED  MECHANICS  FOR  ENGINEERS 

Problem  461.  Find  the  value,  or  values,  of  O  for  steady  pre- 
cession with  axis  inclined  75°  from  the  vertical  of  the  disk  of  Profr 
lera  454,  the  thickness  of  the  disk  being  negligible. 

What  is  the  least  value  of  o>  that  would  satisfy  the  conditions  for 
the  motion  of  this  disk  when  6  =  75°  ? 

Problem  462.  A  conical  top  is  made  of  wood  and  is  spinning 
about  its  axis  with  a  velocity  of  20  revolutions  per  second,  The  cone 
has  a  base  of  2  in.  and  a  height  of  .2  in.,  and  spins  on  the  apex. 
While  spinning  steadily  with  its  axis  vertical  (sleeping),  it  is  dis- 
turbed by  a  blow  so  that  its  axis  is  inclined  at  an  angle  of  30°  with 
the  vertical.  Find  the  velocity  of  precession  and  the  torque  that  tends 
to  keep  the  top  from  falling. 

(5)  Unsteady  Precession.  Assume  0,  ft),  and  H  to  be 
variables.  The  equations  then  become 

CDJ  =  —  O  sin  0,  ft)2  =  a)',  ft>3  =  XI  cos  $, 

hi  =  —  A£l  sin  6,        h2  =  Aw',       h8  =  (7(a>  +  XI  cos  0), 

Ta=-  A(Q,  cos  0.<o'+  sin  02*\-A<o'Q,  cos  0 


Tb  =  A        +  CO,  sin  <9O  +  &  cos  (9)  -  AW  sin  0  cos  0,  (2') 
at 


).  (3') 

at 

Again  suppose  the  weight  and  the  reaction  of  the 
support  to  be  the  only  impressed  forces.  Then  Ta  =  0, 
Tb  =  Wb  sin  0,  Tc  =  0.  Equation  (3')  then  becomes 


at 
Therefore 

ft)  -h  H  cos  0  —  a  constant. 


DYNAMICS   OF  RIGID  BODIES  387 

For  the  sake  of  simplicity  suppose  the  body  to  have 
been  placed  with  its  geometric  axis  at  rest  making  an 
angle  00  with  the  vertical,  the  body  then  given  an  angular 
velocity  &)0  about  its  axis  and  released.  Then  the  value  of 
co  +  ft  cos  6  at  the  start  was  o>0.  Therefore  throughout 
the  motion 

w  -|-  n  cos  o = &>Q. 

Substituting  this  value  in  equation  (!')  where    Ta  =  Q, 

rlfi 

and  replacing  o>'  by  — -,  it  becomes 

Cvt> 

2  ACl  cos  edO  +  A  sin  OdQ,  -  C^dQ  =  0. 
If  this  equation  be  multiplied  through  by  sin  0,  it  can  then 
be  integrated,  for  it  becomes 

A(2Q  sin  0  cos  Bd6  +  sin2  0<W1)  -  (7w0  sin  0d6=  0, 

in  which   the    quantity   inclosed   in   the   parenthesis    is 
d(H  sin2  (9). 
Integrating, 

A£l  sin2  0  -|-  <7o>0  cos  6  =  constant. 

At  the  beginning  of  motion  H  =  0,  and  6  =  00. 
Therefore 

A£l  sin2  0  =  <7ft>0  (cos  00  -  cos  (9).  (5') 

This  equation  expresses  H  in  terms  of  6. 

Instead  of  using  equation  (2')  a  simpler  equation  is  ob- 
tained by  considerations  of  work  and  energy.     The  work 
done  from  the  beginning  of  motion  to  the  instant  under 
consideration  is  that  done  by  gravity,  i.e. 
TT5(cos  <90  -  cos  6). 


388  APPLIED  MECHANICS  FOB  ENGINEERS 

The  gain  in  kinetic  energy  in  this  time  is 

5  !  A(  -  ft  sin  (9)2  +  |  ^W2  +  i  <7a>2j  -  1  (7a>2      (Art.  199) 
Therefore 


w     =          cos    Q     cos 
or  o>'2  =  (CoS  0Q  -  cos  0)  -  ft2  sin2  0. 


Replacing  cos  <90  -  cos  0  by  m2  e  f  rom  equation  (5'), 

(7a>0 

-  n)  sin2  19.  (60 


The  factors  H  and  —  --  H  must  both  be  of  the 


same  sign 


since  a/2  must  be  positive.  Assuming  a>0  to  be  positive,  it 
follows  that  fi  is  positive,  for  if  £t  were  negative,  the  sec- 
ond factor  would  be  positive  and  their  product  would  be 
negative. 

Also  the  values  of  £1  that  make  CD'  zero  are  0  and 


(7o)0 


From  equation  (5') 

o  _  flpp    cos  #Q  —  cos  # 
=  ^1   '        sin2  8 

from  which  it  can  be  shown  that 


Since  sin  0  is  positive,  —  -  is  positive;  i.e.  fl  increases  as 

dO 

0  increases  and  decreases  as  0  decreases.     It  follows  then 
that  0  and  ft  both  increase  until  fl  =  --     &'  is  then  0,  0 


DYNAMICS  OF  RIGID  BODIES  389 

changes  from  increasing  to  decreasing  and  continues  to 
decrease   until   fl  =  0   and   0  =  60   (Equations    (6')    and 

(5'))- 

The  axis  of  the  body  therefore  alternately  falls  and  rises 

between  the  values  6  =  00  and  6  =  Ov  where  6l  is  the  value 

obtained  by  putting  fi  =  — —  in  equation  (5').     As  the 

6o)0 

axis  of  the  body  falls  and  rises,  the  value  of  11  increases 
from  0  when  6  =  6Q  to  — —  when  0  =  0V  decreasing  again 

to  0  when  6  =  00. 

The  condition  for  this  motion  is  that  the  value  of  0  ob- 
tained from  equations  (5')  shall  be  real. 

Problem  463.  Using  the  body  of  Problem  454,  making  a>0  =  600 
r.  p.  m.  and  00  =  30°,  find  the  range  of  values  for  0  and  O. 

Problem  464.  For  what  least  value  of  o>0  would  the  body  of  the 
preceding  problem  have  the  motion  described  in  this  article,  the  other 
conditions  being  the  same  ? 


CHAPTER  XV 
IMPACT 

207.  Definitions.  —  When  two  bodies  collide,  they  are  said 
to  be  subject  to  impact. 

When  the  velocity  of  the  striking  surfaces  is  in  the  direc- 
tion of  the  normal  to  those  surfaces,  the  impact  is  said  to  be 
direct.  Otherwise  it  is  oblique. 

When  the  normal  to  the  surfaces  at  the  point  of  contact 
(or  center  of  forces  exerted  by  one  body  on  the  other) 
passes  through  both  centers  of  gravity  of  the  bodies,  the 
impact  is  said  to  be  central. 

The  phenomena  of  impact  may  be  best  studied  by  con- 
sidering the  two  bodies  somewhat  elastic.  Suppose  for 
simplicity  that  they  are  two  spheres,  Ml  and  M2  (Fig.  280), 
and  that  they  are  moving  with  velocities  v1  and  t»2  and 
that  the  impact  is  central.  In  Fig.  280  (a)  they  are  shown 
at  the  instant  when  contact  first  takes  place,  and  in  Fig.  280 
(5)  they  are  shown  some  time  after  first  contact  when  each 
has  been  deformed  somewhat  by  the  pressure  of  the  other. 
The  dotted  lines  indicate  the  original  spherical  form  and 
the  full  lines  the  assumed  form  of  the  deformed  spheres. 
When  the  spheres  first  touch,  the  pressure  P  between  them 
is  zero,  but  as  each  one  compresses  the  other,  the  pressure 
P  increases  until  it  becomes  a  maximum.  The  compression 
of  the  spheres  is  indicated  in  the  figure  by  dl  and  d2.  We 

390 


IMPACT 


391 


form.      The 
pressure     P 


shall  designate  the  time  during  which  the  bodies  are  being 
compressed  as  the  period  of  deformation. 

After  the  compression  has  reached  its  maximum  value 
the  bodies,  if  they  be  partially  elastic,  begin  to  separate  and 
to  regain  their  original  v 

common 
decreases 

arid  becomes  zero,  if  the 
bodies  are  sufficiently 
elastic  so  that  they 
finally  separate.  We 
shall  designate  this  pe- 
riod of  separation  as  the 
period  of  restitution,  and 
the  velocities  after  sep- 
aration as  v[  and  v%> 

If  the  bodies  are  en- 
tirely inelastic,  there  will 
be  no  restitution.  They 
will,  in  that  case,  remain 


PIG.  280 


in  contact  just  as  they 
are  when  the  pressure  be- 
tween them  is  amaximum 
and  will  move  on  with  a 
common  velocity  V. 

In  what  follows  velocities  and  accelerations  toward  the 
right  will  be  called  positive  and  those  toward  the  left 
negative. 

208.  Direct  Central  Impact.  —  When  the  bodies  meet  in 
direct  central  impact,  separation  will  take  place  along  the 


392  APPLIED  MECHANICS  FOR   ENGINEERS 

line  joining  the  centers  of  gravity.  Let  T  be  the  time 
from  the  first  contact  up  to  the  time  of  maximum 
pressure,  that  is,  the  time  of  deformation,  and  Tt  the  time 
from  first  contact  up  to  the  time  of  separation.  Then 
T^  —  T  represents  the  time  of  restitution.  We  have,  from 
Art.  103,  dv  =  a  •  dt.  Considering  the  motion  of  Ml 
during  the  period  of  deformation,  we  have  dv  =  adt  and 

--k 

(*V  1        (+T 

so  that  1  dv=—±-  I    Pdt, 

•/•i  M^o 

or 

where  J^is  the  common  velocity  of  the  centers  of  gravity 
of  the  bodies  at  the  end  of  the  period  of  deformation. 
In  a  similar  way, 


XT 
Pdt  on  the  right-hand  side  of  the 
_ 

preceding  equations  cannot  be  determined  since  we  do  not 
know  in  general  how  the  pressure  P  varies  with  the  time  ; 
we  do  know,  however,  that  they  are  equal  term  for  term, 
so  that  we  may  eliminate  them.  We  have,  then, 


or  (Mi  4-  Mz)  V  = 

If  the  impact  is  not  too  severe,  elastic  or  partially  elastic 
bodies  tend  to  regain  their  original  shape  after  the  defor- 
mation has  reached  a  maximum  and  finally  separate  if  they 


IMPACT  393 

possess  sufficient  elasticity.  Using  the  notation  of  Art. 
207,  we  have,  for  the  period  of  restitution,  if  E  is  the  force 
of  restitution, 

dv=- 


so  that  Ml(y!L-V)=- 

and 

from  which       (Ml  +  M^)  V  =  M^  + 

The  value  of  the  integral  j  Pdt  during  deformation  will 
not  in  general  be  the  same  as  its  value  during  restitution. 

Call  the  ratio  of  C  *Bdt  to  f  Pdt,  e.     This  value,  which 
JT  */o 

is  called  the  coefficient  of  restitution,  is  constant  for  given 
materials.  It  is  unity  for  perfectly  elastic  substances,  zero 
for  non-elastic  substances,  and  some  intermediate  value 
for  the  imperfectly  elastic  materials  with  which  the  en- 
gineer is  usually  concerned.  The  following  values  of  e 
have  been  determined  :  for  steel  on  steel,  e  =  .55  ;  for  cast 
iron  on  cast  iron,  e  =  1,  nearly  ;  for  wood,  e  =  0,  nearly. 

From  the  above  definition  of  e,  it  is  seen  at  once  that  we 
may  write 

v'-V 


V—  vl 


,  and 


from  which  it  follows  that 


394  APPLIED  MECHANICS  FOR  ENGINEERS 


where 


From  the  above  equations  it  is  seen  that 


209.  Momentum  and  Kinetic  Energy  in  Impact  __  When  a 
constant  force  acts  upon  a  body  for  a  certain  time,  the 
change  in  the  velocity  of  the  body  is  directly  proportional 
to  the  force  and  to  the  time  and  inversely  proportional  to 
the  mass  of  the  body. 

The  product  of  the  force,  F,  and  the  time,  t,  is  called 
the  impulse  of  the  force,  and  the  product  of  the  mass  and 
velocity  is  called  the  momentum  of  the  body.  (The  body 
is  supposed  to  have  no  rotation.) 

From  Newton's  law,  F  =  Ma,  it  follows  that 

Ft  =  M  (*!  -  t>0), 

or,  when  a  body  is  acted  upon  ly  a  constant  force,  the  im- 
pulse of  the  force  is  equal  to  the  change  in  momentum  of  the 
body. 

If  the  force  is  not  constant,  and  is  represented  at  any 

time  by  P,  then   I   Pdt  is  the  impulse  of  the  force   for 

*/o 

the  time  from  0  to  t,  and  from  the  equations  of  the  preced- 
ing article, 


f 

Jo 


IMPACT  395 

it  follows   that  the  impulse  of  the  force  is  equal  to  the 
change  of  the  momentum  of  the  body  on  which  it  acts,  no 
other  force  being  assumed  to  act  on  the  body. 
From  the  equations  of  the  preceding  article, 


and  jtf  +  M)  V= 


it  follows  that  there  is  no  change  in  the  sum  of  the  mo- 
menta of  the  bodies  in  impact,  either  during  the  period 
of  deformation  or  the  period  of  restitution.  This  is 
otherwise  evident  from  the  fact  that  the  forces  acting  on 
the  bodies  are  equal  and  opposite,  and  hence  what  one 
body  gains  in  momentum  the  other  loses. 

With  the  kinetic  energy  it  is  different.      During  the 
period  of  deformation  the  loss  in  kinetic  energy  is 

El  =  I  My\ 


During  the  period  of  restitution  there  is  a  gain  in  kinetic 

,    ,2 
3 

,2 

l" 


since  v{  —  v^  —  —  e(yl  —  v^). 

The  total  loss  in  kinetic  energy  during  the  impact  is  there- 

f  ore  F  -  F  =  M^Mzd- 

2(  jfi 


396  APPLIED  MECHANICS  FOR  ENGINEERS 

If  the  bodies  are  inelastic,  e  =  0,  and  the  bodies  go  on 
with  the  common  velocity  F".  The  loss  in  kinetic  energy 
in  this  case  is 


If  the  bodies  are  perfectly  elastic,  e  =  1,  and  the  loss  in 
kinetic  energy  is  zero. 

Problem  465.  A  lead  sphere  whose  radius  is  2  in.  strikes  a  large 
mass  of  cast  iron  after  falling  freely  from  rest  through  a  distance 
of  100ft.  What  is  its  final  velocity?  What  is  the  loss  of  kinetic 
energy?  Assume  e  =  0.  Lead  weighs  710  Ib.  per  cubic  foot. 

Problem  466.  A  10-lb.  lead  sphere  is  at  rest  when  it  is  acted 
upon  by  another  lead  sphere,  whose  radius  is  3  in.,  in  direct  central 
impact.  The  velocity  of  the  latter  sphere  is  20  ft.  per  second.  What 
is  the  common  velocity  of  the  two  spheres  and  what  is  the  loss  of 
kinetic  energy  due  to  impact  ? 

Problem  467.  Prove  that  if  two  inelastic  bodies,  moving  in 
opposite  directions  with  speeds  inversely  proportional  to  their  masses, 
collide,  both  will  be  reduced  to  rest. 

Problem  468.  Prove  that  if  two  perfectly  elastic  bodies  with 
equal  masses  collide,  each  mass  will  have  after  impact  the  velocity 
that  the  other  had  before.  If  one  of  the  bodies  is  at  rest  and  the 
other  strikes  it,  what  will  happen  ? 

Problem  469.     Figure  281  represents  a   set   of   perfectly  elastic 

-  »•  balls  of  equal  mass  and  size  in  contact 

-  -  Q  -  OOOOOO  -    in  a  straight  line.     Another  ball,  of 


the  same  mass  and  material,  moving 
in  the  direction  of  the  line  of  balls, 

— >  strikes  one  end  of  the  line.      Prove 

CO         OOOOOO —         that  the  moving  ball  will  be  brought 


FlG-  281  to  rest  where  it  strikes  and  the  last 

ball  at  the  other  end  of  the  line  will  take  its  velocity,  all  other  balls 
remaining  at  rest.      Prove  also  that  if  two  balls,  moving  together, 


IMPACT  397 

strike  the  line,  the  two  balls  at  the  other  end  of  the  line  will  take  the 
velocity  of  the  striking  balls  and  the  latter  will  be  brought  to  rest. 

Problem  470.  Prove  that  if  a  ball  of  mass  M\  fall  from  a  height 
H  upon  a  very  large  mass  with  a  horizontal  surface  and  rebounds  to 
a  height  h,  the  coefficient  of  restitution  of  the  materials  used  is 


^H 

(Regard  M2  as  infinite  and  V  =  0.) 

Problem  471.  A  bullet  weighing  1  oz.  is  fired  horizontally  into  a 
box  of  sand  (inelastic)  weighing  20  lb.,  and  remains  imbedded  in  the 
sand.  The  box  is  suspended  by  a  string  attached  to  a  fixed  point 
4  ft.  from  the  center  of  the  box  of  sand.  The  impact  of  the  bullet 
causes  the  box  to  swing  aside  through  an  angle  of  42°.  Find  (a)  the 
velocity  of  impact  of  the  bullet,  (b)  the  kinetic  energy  lost  in  impact, 
(c)  the  greatest  tension  in  the  string. 

Problem  472.  Assuming  inelastic  impact,  show  that  if  a  pile 
driver  weighing  Gl  lb.  falls  h  ft.  and  drives  a  pile  weighing  G2  lb.  s  ft. 
into  the  ground  against  a  constant  resistance  R, 


R= 


Gzs 


(HiNT.  Equate  work  done  against  resistance  plus  kinetic  energy 
lost  in  impact  to  the  work  done  by  gravity  on  the  falling  weight.) 

Problem  473.  A  wooden  pile  weighing  1500  lb.  is  driven  by  a 
steel  hammer,  of  weight  2000  lb.,  falling  20  ft.  The  penetration  at 
the  last  blow  is  observed  to  be  \  in.  What  is  the  resistance  offered 
by  the  earth  to  the  pile  ?  With  a  safety  factor  6,  what  load  would 
the  pile  carry  ? 

210.  Elasticity  of  Material.  —  All  materials  of  engineer- 
ing are  imperfectly  elastic.  Some,  however,  show  almost 
perfect  elasticity  for  stresses  that  are  rather  low.  This 
has  been  expressed  by  saying  that  all  materials  have  a 


398  APPLIED  MECHANICS  FOR  ENGINEERS 

limit  (elastic  limit)  beyond  which  if  the  stress  be  increased 
the  material  will  be  imperfectly  elastic.  Within  the 
limit  of  elasticity,  stress  is  proportional  to  the  deforma- 
tion produced.  Let  P  be  the  total  stress  in  tension  or 
compression,  /  the  stress,  in  pounds  per  square  inch  of 
cross  section,  d  be  the  deformation  caused  by  P  and  X 
the  deformation  per  inch  of  length.  Within  the  limit 

of  elasticity  of  the  material  the  ratio  2-  is  a  constant,  and 

p  d  x 

since/  =  —  ,  and  X  =  -,  when  F  is  the  area  of  cross  sec- 

F  I 

tion  and  I  is  the  length  of  the  material,  it  may  be  written 

PI 

—  •  This  constant  is  called  the  modulus  of  elasticity  of 
Fd 

the  material;  it  is  usually  represented  by  E,  so  that 


\     Fd 

for  tension  or  compression.     For  steel  E  has  been  found 
to  be  about  30,000,000  Ib.  per  square  inch. 

211.  Impact  Tension  and  Impact  Compression.  —  Figure 
282  (V)  represents  a  mass  M2  subjected  to  impact  from 
the  mass  Ml  falling  from  rest  through  a  height  h.  The 
mass  Mz  is  compressed  by  the  impact.  Figure  282  (i) 
represents  the  body  M2  as  subjected  to  impact  in  tension, 
the  mass  in  this  case  being  a  rod  having  M1  attached  to  one 
end  and  the  other  end  attached  to  a  crosshead  A.  The 
rod,  crosshead,  and  weight  fall  freely  together  through 
the  distance  h  until  A  strikes  the  stops  at  B,  when  one  end 
of  the  rod  suddenly  comes  to  rest  and  the  mass  Ml  causes 
tension  in  the  rod  due  to  impact. 


IMPACT 


399 


Suppose  vl  represents  the  velocity  of  Ml  when  impact 
occurs  and  Z2  the  length  of  M^  whether  it  be  a  tension  or 
compression  piece. 


FIG.  282 


In  case  (5)  when  the  weights  strike  the  stops  at  B  the 
kinetic  energy  of  the  falling  weights  is  equal  to 


where  6rx  and  6r2  are  the  weights  of  Ml  and  M%  in  pounds 
and  h  is  in  feet.  This  kinetic  energy  is  used  up  in  doing 
work  in  compressing  the  stops  and  in  stretching  Ml  and 
M2.  If  M2  is  a  rod  of  small  cross  section,  its  length  is  in- 
creased a  large  amount  compared  to  the  change  in  length 
of  M1  and  the  stops  against  which  the  crosshead  strikes, 


400  APPLIED   MECHANICS  FOR  ENGINEERS 

and  we  may  without  serious  error  assume  that  all  the  ki- 
netic energy  is  used  up  in  stretching  M2.  We  may  assume 
too  that  the  stress  in  M2  is  the  same  throughout  its  length. 

Let  Pm  be  the  maximum  stress  induced  in  the  rod.    The 

p 
average  stress  is  then  — -,  and  if  dm  is  the  total  increase  in 

2 

the  length  of  the  rod,  the  work  done  on  the  rod  is 


But  pm  =  Ja«m^a.  (Art.  210) 


Here  E  must  be  expressed  in  pounds  per  square  foot  and 
all  dimensions  in  feet  if  Crl  and  6r2  are  in  pounds  and  h  in 
feet. 

In  case  (a)  the  student  can  easily  show  that 


Problem  474.     Derive  the  formula  just  given  for  the  case  (a). 

Problem  475.  A  weight  of  500  Ib.  falls  through  a  distance  of  2  ft. 
in  such  a  way  as  to  put  a  1-in.  round  steel  rod  in  tension.  If  the  rod 
is  18  ft.  long,  what  will  be  the  elongation  due  to  the  impact?  Use 
E  =  30,000,000  Ib.  per  square  inch  for  steel.  What  maximum  unit 
stress  is  induced  in  the  rod  ? 


IMPACT  401 

Problem  476.  A  cylindrical  piece  of  steel  1  in.  high  and  1  in.  in 
diameter  is  subjected  to  compression  by  a  weight, of  20  Ib.  falling 
through  a  distance  of  1  in.  How  much  will  it  be  compressed? 

Problem  477.  In  the  preceding  problem  what  stress  (pounds  per 
square  inch)  was  caused  in  the  cylindrical  block  by  the  fall  of  the 
20-lb.  weight? 

Problem  478.  The  safe  stress  in  structural  steel  for  moving  loads 
is  usually  taken  as  12,500  Ib.  per  square  inch.  Through  what  height 
might  a  300-lb.  weight  fall  so  as  to  produce  tension  in  a  1-in.  steel 
round  bar,  10  ft.  long,  without  exceeding  the  safe  stress? 

Problem  479.  Two  steel  tension  rods  in  a  bridge,  each  2  sq.  in.  in 
cross  section  and  20  ft.  long,  carry  the  effect  of  the  impact  of  a  loaded 
wagon  as  one  wheel  rolls  over  a  stone  1  in.  high.  The  weight  on  the 
wheel  is  2000  Ib.  What  stress  is  introduced  in  the  tension  rods  ? 

NOTE.  For  the  strength  of  metals  under  impact  the  student  is 
referred  to  the  work  of  W.  K.  Hatt,  Am.  Soc.,  «  Testing  Materials," 
Vol.  IV,  p.  282. 

212.   Direct  Eccentric  Impact.  —  Let   Ml   and   M2  have 
impact   as   shown   in  Fig.  283,  in  which   the  centers  of 
gravity  of  the  bodies  are  moving  along 
parallel  lines  and  the  surfaces  of  contact 
are  perpendicular  to  the  line  of  motion. 
This  is  known  as  direct  eccentric  impact.      

Let  the  velocity  of  M%  before  impact 
be  Vy,  the  velocity  of  the  center  of  gravity 
of  MI  be  Vj,  and  its  angular  velocity  be 


6 

LJ 


(a)  The  problem  will  first  be  solved 
on  the  assumption  that  M-±  is  acted  upon 
by  no  forces  except  that  of  impact. 

First  period  of  impact.     Let  P  be  the  force  at  any  in- 

2D 


402  APPLIED  MECHANICS  FOR  ENGINEERS 

stant  of  the  first  period  of  impact,  T  the  time  of  the  first 
period,  Vthe  Velocity  of  Mv  &>  the  angular  velocity  of  Mr 
and  Fj  the  velocity  of  the  center  of  gravity  of  Ml  at  the 
end  of  the  first  period. 

During  the  first  period,  which  lasts  for  a  very  short 
time,  the  position  of  the  body  M1  may  be  regarded  as 
unchanged.  We  then  have  for  the  motion  of  M^ 


va),  (1) 

CTPdt  =  Ml  f  Fl  dv  =  JKi(  Fj  -  ^),  (2) 

*/0  Jvi 


and  for  the  motion  of  Ml 


and  bP  =  l 

dt 


or 


>  CTPdt  =  JO2  C^dco  =  MJc\(*>  -  &),),  (3) 

*/o  */Wl 

where  k  is  the  radius  of  gyration  of  M^  about  a  gravity 
axis  perpendicular  to  the  plane  of  motion.  Also,  since  at 
the  end  of  the  first  period  the  velocity  of  M2  is  the  same 
as  the  velocity  of  the  point  of  impact  on  Mv 

The  unknowns  in  these  four  equations  are 

*T 

Pdt,  F;  rr  o>. 

The  equations  are  linear  in  these  unknowns  and  the  un- 
knowns can  be  easily  determined  from  them. 

Second  period  of  impact.  If  R  is  the  force  of  impact  at 
any  time  during  the  second  period  of  impact,  T  the  value 
of  £,  #2  the  velocity  of  M^  v[  the  velocity  of  the  center  of 


IMPACT 


403 


gravity  of  Mv  and  a>{  the  angular  velocity  of  M1  at  the  end 
of  the  second  period,  then 


-        Edt 


z  dv  =  Jf2<X  -  F), 


dv  = 


(5) 
(6) 


Also 


i'  -  a)).         (7) 

(8) 


X. 


Here   again   are  four   linear   equations   in   the  four  un- 
knowns, I     Rdt,  vL  vL  ft)/, 
JT 

which  are  therefore  sufficient  to  determine  the  unknowns. 

(6)  Impact  on  a  body  with  fixed  axis.  Suppose  the  body 
rotating  about  the  fixed  axis  through  0  (Fig.  284),  with 
angular  acceleration  a.^  and  angular 
velocity  o^  at  the  beginning  of  im- 
pact. 

Choose  the  origin  of  coordinates 
at  0  and  the  #-axis  coinciding  with 
the  direction  of  motion  of  the  cen- 
ters of  gravity  of  the  bodies. 

Let  X  and  Y  be  the  reactions  of 
the  supporting  axis  through  0  at 
any  time  during  the  impact. 

First  period  of  impact.  For  the 
first  period  we  have  the  following  equations : 


7 

c 

.    0 

1: 

hr 

X 

FIG.  284 


404  APPLIED  MECHANICS  FOK  ENGINEERS 


X)dt  =  JMi(  Vl  -  t^),  (2') 


Also,  since  0  is  fixed, 

Fi  =  (A-J)o>,  (4') 

F=  A®.  (5') 

Here  are  five  equations  containing  five  unknowns: 

CTpdt,    Cxdt,    vv    F;   a,, 

»/  »/o 


from  which  the  unknowns  may  be  determined. 

Second  period  of  impact.         For  the  second  period  the 
following  equations  hold  : 


=  M£v'i-  F),  (6') 

^MM-V^  (70 

-  X(h  -  iy\dt  =  MJPM  -  to),  (80 

(90 


Here  again  are  five  equations  containing  five  unknowns  : 

C^Bdt,      CT>Xdt,     vd     v'v  co{. 
JT  JT 

These  equations  are  sufficient  to  determine  the  unknowns, 
and  thus  the  motion  of  the  bodies  is  given  after  impact. 

In  the  above  equations  k  is  the  radius  of  gyration  of  Ml 
about  a  gravity  axis  perpendicular  to  the  plane  of  motion. 


IMPACT  405 

Problem  480.  A  uniform  steel  bar,  4  ft.  long,  weighing  16  lb., 
lies  at  rest  on  a  smooth  horizontal  plane.  It  is  struck  by  a  steel  ball 
weighing  2  lb.,  with  a  velocity  of  30  f/s,  at  right  angles  to  the  rod  at 
a  point  6  in.  from  one  end.  Find  the  velocity  of  the  ball,  the  velocity 
of  the  center  of  gravity  of  the  bar,  and  its  angular  velocity  at  the  end 
of  the  first  period,  and  at  the  end  of  the  second  period  of  impact, 
given  e  =  .55. 

Problem  481.  In  the  preceding  problem  find  the  point  about 
which  the  bar  is  turning  at  the  end  of  the  first  period.  Is  this  point 
the  center  of  rotation  at  the  end  of  the  second  period  ? 

Problem  482.  If  Mi  is  at  rest  when  M2  strikes  with  direct  eccen- 
tric impact,  write  the  equations  which  determine  the  motion  at  the 
end  of  the  first,  and  at  the  end  of  the  second  period  of  impact. 

Problem  483.  If  e  =  0  between  MI  and  M2  of  Art.  212  (a),  find 
the  kinetic  energy  lost  in  impact. 

Problem  484.  Suppose  MI  to  be  a  rod  of  steel  |  in.  in  diameter 
and  2  ft.  long,  and  suppose  M2  to  be  a  hammer  weighing  2  lb.  and 
that  its  velocity  at  the  time  of  impact  is  20  ft.  per  second,  find  V  and 
<u,  if  the  hammer  strikes  10  in.  from  the  center  of  the  rod.  e  =  .55. 

Problem  485.  Let  MI  be  a  square  stick  of  timber  4"  x  4"  x  10' 
and  let  M2  be  a  10-lb.  hammer  having  a  velocity  at  the  time  of  impact 
of  10  ft.  per  second.  If  the  impact  takes  place  4  ft.  from  the  center, 
find  V  and  <o,  given  e  =  .10, 

213.  Center  of  Percussion  and  Center  of  Instantaneous  Rota- 
tion. —  Consider  a  body  at  rest  to  be  struck  a  blow  and  to 
be  free  to  move  in  the  plane  passing  through  the  center 
of  gravity  of  the  body  and  the  line  in  which  the  blow  is 
struck;  as,  for  example,  a  slab  at  rest  on  a  smooth  hori- 
zontal plane  when  struck  a  blow  in  a  horizontal  plane 
through  the  center  of  gravity  of  the  body. 

Let  0  be  the  center  of  gravity  of  the  body  and  P  the 
force  of  the  blow  at  any  time  during  the  impact  (Fig. 


406 


APPLIED  MECHANICS  FOR  ENGINEERS 


285).     Then,  since  for  the  very  brief  period  in  which  the 
blow  takes  place  any  change  in  the  position  of  the  body 

is  negligible,  we  have  for  the 
motion  of  the  body  at  any  in- 
stant during  the  blow,  if  v  is  the 
velocity  of  the  center  of  gravity, 
and  ft>  the  angular  velocity, 

*-«»• 


aa 

A 

V 

v# 

't?/^\ 

i 

1 

N.7* 

\ 

.  V 

c 

5 

where  &  is  the  radius  of  gyration 

about  a  gravity  axis  perpendicular  to  the  plane  of  motion 
of  the  center  of  gravity  of  the  body. 

If  t  is  the  time  at  any  instant  of  the  blow,  measured 
from  the  beginning  of  the  blow,  there  follow  from  the 
above  equations, 


and  b  Cpdt  =  Mk2  C d<*  =  Mk2co. 

»/o  »/o 

By  division,  j.~T*~' 

J  b      k  co 

The  velocity  of  any  point  of  the  body  at  the  given 
instant  is  composed  of  two  components,  one  the  velocity, 
v,  of  the  center  of  gravity,  and  the  other,  rco,  relative  to 
the  center  of  gravity.  There  is  then  one  point  of  the 
body,  or  in  a  plane  fixed  in  the  body  and  moving  with  it, 
where  these  two  components  just  annul  each  other.  This 
point  (^.,  Fig.  285)  must  lie  on  the  line  through  the 


IMPACT  407 

center  of  gravity  perpendicular  to  the  line  of  the  blow  at 
a  distance  a  from  the  center  of  gravity  such  that 

aco  =  v. 

But  7  =  if' 

b      k2co 

By  division,  ab  =  K2, 

from  which  a  is  determined  when  b  and  k  are  known. 

There  is,  therefore,  an  axis  for  each  blow  about  which 
the  body  begins  to  turn,  and  this  axis  remains  approxi- 
mately the  same  as  long  as  the  change  in  the  position  of  the 
body  is  negligible ;  that  is,  in  general,  during  the  blow. 

The  axis  about  which  the  body  begins  to  turn  under 
the  action  of  the  blow  is  called  the  instantaneous  axis  of 
rotation. 

The  point,  A,  in  this  axis  which  lies  in  the  plane  con- 
taining the  center  of  gravity  and  the  line  of  the  blow  is 
called  the  instantaneous  center  of  rotation. 

If  a  fixed  axis  passed  through  the  body  at  A,  perpen- 
dicular to  the  plane  containing  the  center  of  gravity  and 
the  line  of  the  blow,  the  blow  would  cause  the  body  to 
turn  about  this  axis  without  causing  any  sudden  reaction 
of  the  axis  on  the  body.  The  point  jB,  the  foot  of  the  per- 
pendicular from  the  center  of  gravity  to  the  line  of  the 
blow,  is  known  in  this  case  as  the  center  of  percussion  of 
the  body  corresponding  to  the  center  of  rotation  A. 

It  should  be  noted  that  center  of  percussion  and  center 
of  instantaneous  rotation  are  related  just  as  center  of 
suspension  and  center  of  oscillation  are  in  the  compound 
pendulum  (Art.  182). 


408 


APPLIED  MECHANICS  FOR   ENGINEERS 


Problem  486.  A  thin  rod,  .3  ft.  long,  is  struck  a  blow  at  6  in. 
from  one  end  and  at  right  angles  to  the  rod.  Find  the  instantaneous 
center  of  rotation. 

Problem  487.  A  right  circular  steel  cylinder  of  diameter  6  in. 
and  height  2  ft.  is  suspended  by  an  axis  in  a  diameter  of  one  end. 
It  is  struck  a  blow  with  a  5-lb.  hammer  with  a  velocity  of  20f/s 
through  the  center  of  percussion  corresponding  to  the  given  support. 
Find  the  velocity  of  the  hammer  and  the  angular  velocity  of  the 
cylinder  at  the  time  of  greatest  pressure  and  at  the  end  of  the  impact, 
given  e  =  .55. 

Problem  488.  A  right  circular  cone  of  steel,  the  radius  of  whose 
base  is  6  in.  and  altitude  6  in.,  is  supported  as  a  pendulum  by  an  axis 
through  its  vertex  parallel  to  the  base.  It  is  struck  with  a  3-lb. 
hammer  with  a  velocity  of  10  ft.  per  second,  in  a  line  through  the 
center  of  percussion.  Find  V  and  o>  at  time  of  greatest  pressure. 

Problem  489.  A  man  strikes  a  blow  with  a  steel  rod  1|  in.  in 
diameter  and  4  ft.  long,  by  holding  the  rod  in  the  hand  and  striking 
the  farther  end  against  a  stone  in  such  a  way  as  to  cause  the  rod  to 
be  under  flexure.  Where  should  he  grasp  the  rod  in  order  that  he 
may  receive  no  shock  ? 

Problem  490.  Prove  that  after  the  blow  has  ceased  to  act  on  the 
body  (Fig.  285)  the  body  would  then  move  in  such  a  way  that  a  circle 
of  the  body  with  radius  CA  and  center  C  would  roll  along  a  straight 
line.  Hence  show  that  every  point  of  the  body  in  this  circle  would 
generate  a  cycloid.  It  is  assumed  that  no  other  forces  act  on  the  body. 

214.  Oblique  Impact  of  Body  against  Smooth  Plane.  —  Let 

M (Fig.  286)  be  a  sphere 
moving  toward  the  plane 
indicated  with  a  velocity 
at  impact  of  w,  the  direc- 
tion of  motion  making  an 
angle  a  with  the  vertical 
to  the  plane.  After  im- 


D 


A 
FIG.  286 


IMPACT  409 

pact  the  body  M  rebounds  with  a  velocity  v1  in  the  direc- 
tion making  an  angle  0  with  the  vertical  AB.  Since 
the  plane  is  considered  smooth,  the  effect  of  the  impact 
will  be  all  in  the  direction  of  AB,  and  hence  the  component 
of  the  velocity  parallel  to  the  plane  will  not  change. 

.  •.  v1  sin  ft  =  v  sin  a.  (1) 

Assuming  that  the  plane  does  not  move,  V=  0,  where  V 
is  the  component  of  the  velocity  of  M  perpendicular  to  the 
plane  at  the  time  of  greatest  pressure.  Then 

rP<fc-lf[0-<>coea)], 
, 

and  f  Jfcft  =  Jf(  -  vl  cos  ft  -  0). 

JT 

Therefore 


X77 
Rdt 


_j  _  ,_  __  e 

vcosa 


or  vl  cos  ft  =  ev  cos  «.  (2) 

Dividing  (1)  by  (2), 

tan  /3  =  -  tan  a. 
e 

Squaring  and  adding  (1)  and  (2), 


Hence  if  a,  v,  and  e  are  known,  vl  and  y8  may  be  deter- 
mined. 

If   the   bodies   are  perfectly  elastic,  e  =  1,  ft  =  a,    and 

z>x  =  v.     If  the  body  is  inelastic,  e  =  0,  j3  =  —  ,  vx  =  v  sin  a. 


410 


APPLIED  MECHANICS  FOR  ENGINEERS 


The  body  then  moves  along  the  plane  with  a  velocity 
v  sin  a. 

Problem  491.  A  ball  is  projected  with  velocity  50  f/s  against  a 
smooth  plane  surface  at  an  angle  with  the  normal  of  40°.  It  leaves  at  an 
angle  of  50°  with  the  normal.  Find  e  and  the  velocity  at  which  it  leaves. 

215.  Impact  of  Rotating  Bodies.  —  Suppose  two  bodies 
Ml  and  Mz  revolve  about  two  parallel  axes  Ol  and  02  (Fig. 

287)  in  such  a  way 
that  impact  occurs  at 
a  point  along  the  line 
DE.  Let  the  line 
along  which  impact 
occurs  be  distant  r1 
and  r2  respectively 
from  Oj  and  02.  Let 
P  be  the  force  of  im- 
pact at  any  instant 
during  the  first  period 
and  R  the  force  at  any  instant  during  the  second  period, 
CDI  and  6>2  the  angular  velocities  at  the  beginning,  <o[ 
and  0)%'  the  angular  velocities  at  the  end  of  impact,  I± 
and  .Zg  the  moments  of  inertia  of  the  bodies  respectively 
about  01  and  02,  and  F'the  rectangular  component  of  the 
velocity  of  the  point  of  impact  in  the  direction  of  DE  at 
the  time  of  greatest  pressure.  The  angular  velocities  of 

the  bodies  at  this  instant  are  then  —  and  —  • 

r*          r» 


FIG.  287 


Then,  as  in  Art.  212, 


IMPACT  411 

(2) 

v/2  / 

f*TI  /•«,*  /  TTX 

-JSdt-lJrd*  =/!(*,;-£),  (3) 

(4) 

,r-  „  (5) 

From  (1)  and  (2), 


From  (1),  (3),  and  (5), 
F         V 


or  o>i  =  —(1  +  e)  —  ewj. 

ri 

Similarly,  ft)'2  =  _(1  +  e)  —  ea)2. 


(It  should  be  noted  that  in  these  formulae  the  angular 
velocities  of  the  bodies  are  reckoned  in  opposite  direc- 
tions.) 

Problem  492.  If  71=  3000,  72  =  15,000,  coj  =  1  radian  per  second, 
o)2  =  0,  rj  =  2  ft.,  r2  =  3  ft.,  and  c  =  0,  find  F,  o>{,  o>^  and  the  kinetic 
energy  lost  in  impact. 


412 


APPLIED  MECHANICS  FOR  ENGINEERS 


Problem  493.  A  well  drill  is  shown  in  principle  in  Fig.  288. 
The  drill  is  supported  by  a  cable  that  passes  over  a  pulley  C  and  is 
attached  to  a  friction  drum  A.  When  A  is  held,  the  drill  is  raised 
by  the  operation  of  M^  and  M2.  Suppose  that  I\  is  300  and  c^  =  3 


FIG.  288 


radians  per  second ;  72  =  200  and  o>2  —  0 ;  rl  =  2  ft.  and  r2  =  6  ft. 
Assume  e  =  \.  Find  <D(  and  o>2'.  What  kinetic  energy  is  lost  due  to 
each  impact  ? 


IMPACT 


413 


Problem  494.    The  moment  of  inertia  of  the  trip   hammer  M2, 
illustrated  in  principle  in  Fig.  289,  is  100,000 ;  that  of  Mx  is  60,000. 


FIG.  289 


If  rx  =  3  ft.,  r2  =  10  ft.,  (D!  =  2  radians  pe'r  second,  wa  =  0,  and  e  =  £, 
find  o>{  and  (Og.  What  is  the  kinetic  energy  lost  due  to  each  impact? 
What  is  the  kinetic  energy  of  the  hammer  ? 


APPENDIX  I 
HYPERBOLIC   FUNCTIONS 


cosh  x  = 

px p-x 

•         1  Cx       ^^   tx 

sinh  x  = 


tanh  x  = 


2 
sinh  a;       e*  —  e~x 


cosh  x      ex  +  e~* 


HYPERBOLIC  FUNCTIONS 


417 


X 

Cosh  a; 

Sinhz 

X 

Cosh  x- 

Sinhx 

0.01 

1.0000500 

0.0100002 

0.51 

1.1328934 

0.5323978 

.02 

.0002000 

.0200013 

.52 

.1382741 

.5437536 

.03 

.0004500 

.0300045 

.53 

.1437686 

.5551637 

.04 

.0008000 

.0400107 

.54 

.1493776 

.5666292 

.05 

.0012503 

.0500208 

.55 

.1551014 

.5781516 

.06 

.0018006 

.0600360 

.56 

.1609408 

.5897317 

.07 

.0024510 

.0700572 

.57 

.1668962 

.6013708 

.08 

.0032017 

.0800854 

.58 

.1729685 

.6130701 

.09 

.0040527 

.0901215 

.59 

.1791579 

.6248305 

.10 

.0050042 

.1001668 

.60 

.1854652 

.6366536 

.11 

.0060561 

.1102220 

.61 

.1918912 

.6485402 

.12 

.0072086 

.1202882 

.62 

.1984363 

.6604917 

.13 

.0084618 

.1303664 

.63 

.2051013 

.6725093 

.14 

.0098161 

.1404578 

.64 

.2118867 

.6845942 

.15 

.0112711 

.1505631 

.65 

.2187933 

.6967475 

.16 

.0128274 

.1606835 

.66 

.2258219 

.7089704 

.17 

.0144849 

.1708200 

.67 

.2329730 

.7212643 

.18 

.0162438 

.1809735 

.68 

.2402474 

.7336303 

.19 

.0181044 

.1911452 

.69 

.2476458 

.7460697 

.20 

.0200668 

.2013360 

.70 

.2551690 

.7585837 

.21 

.0221311 

.2115469 

.71 

.2628178 

.7711735 

.22 

.0242977 

.2217790 

.72. 

.2705927 

.7838405 

.23 

.0265668 

.2320333 

.73 

.2784948 

.7965858 

.24 

.0289384 

.2423107 

.74 

.2865248 

.8094107 

.25 

.0314132 

.2526122 

.75 

.2946833 

.8223167 

.26 

.0339908 

.2629393 

.76 

.3029713 

.8353049 

.27 

.0366720 

.2732925 

.77 

.3113896 

.8483766 

.28 

.0394568 

.2836731 

.78 

.3199392 

.8615330 

.29 

.0423456 

.2940819 

.79 

.3286206 

.8747758 

.30 

.0453385 

.3045203 

.80 

.3374349 

.8881060 

.31 

.0484361 

.3149891 

.81 

.3463831 

.9015249 

.32 

.0516384 

.3254894 

.82 

.3554658 

.9150342 

.33 

.0549460 

.3360222 

.83 

.3646840 

.9286347 

.34 

.0583590 

.3465886 

.34 

.3740388 

.9423282 

.35 

.0618778 

.3571898 

.85 

.3835309 

.9561160 

.36 

.0655029 

.3678265 

.86 

.3931614 

.9699993 

.37 

.0692345 

.3785001 

.87 

.4029312 

.9839796 

.38 

.0730730 

.3892116 

.88 

.4128413 

0.9980584 

.39 

.0770189 

.3999619 

.89 

.4228927 

1.0122369 

.40 

.0810724 

.4107523 

.90 

.4330864 

.0265167 

.41 

.0852341 

.4215838 

.91 

.4434234 

.0408991 

.42 

.0895042 

.4324574 

.92 

.4539048 

.0553856 

.43 

.0938888 

.4433742 

.93 

.4645315 

.0699777 

.44 

.0983718 

.4543354 

.94 

.4753046 

.0846768 

.45 

.1029702 

.4653420 

.95 

.4862254 

.0994843 

.46 

.1076788 

.4763952 

.9(5 

.4972947 

.1144018 

.47 

.1124983 

.4874959 

.97 

.5085137 

.1294307 

.48 

.1174289 

.4986455 

.98 

.5198837 

.1445726 

.49 

.1224712 

.5098450 

.99 

.5314057 

.1598288 

0.50 

1.1276260 

0.5210953 

1.00 

1.5430806 

1.1752012 

418 


HYPERBOLIC  FUNCTIONS 


X 

Coshx 

Sinhz 

X 

Coshx 

Sinhx 

1.01 

1.5549100 

1.1906910 

1.51 

2.3738201 

2.1529104 

1.02 

.5668948 

.2062999 

1.52 

.3954676 

.1767566 

1.03 

.5790365 

.2220294 

1.53 

.4173563 

.2008206 

1.04 

.5913358 

.2378812 

1.54 

.4394857 

.2251046 

1.05 

.6037945 

.2538567 

1.55 

.4618591 

.2496111 

1.06 

.6164134 

.2699576 

1.56 

.4844787 

.2743426 

1.07 

.6291940 

.2861855 

1.57 

.5073467 

.2993014 

1.08 

.6421375 

.3025420 

1.58 

.5304654 

.3244903 

1.09 

.6552453 

.3190288 

1.59 

.5538373 

.3499117 

1.10 

.6685186 

.3356474 

1.60 

.5774645 

.3755679 

1.11 

.6819587 

.3523997 

1.61 

.6013494 

.4014618 

1.12 

.7005670 

.3642872 

1.62 

.6254945 

.4275958 

1.13 

.7093449 

.3863116 

1.63 

.6499021 

.4539726 

1.14 

.7232938 

.4034746 

1.64 

.6745748 

.4805947 

1.15 

.7374148 

.4207781 

1.65 

.6995149 

.5074650 

1.16 

.7517098 

.4382235 

1.66 

.7247249 

.5345859 

1.17 

.7661798 

.4558128 

1.67 

.7502074 

.5619603 

1.18 

.7808265 

.4735477 

1.68 

.7759650 

.5895910 

1.19 

.7956513 

.4914299 

1.69 

.8020001 

.6174806 

1.20 

.8106556 

.5094613 

1.70 

.8283154 

.6456319 

1.21 

.8258410 

.5276436 

1.71 

.8549136 

.6740479 

1.22 

.8412089 

.5459788 

1.72 

.8817974 

.7027311 

1.23 

.8567610 

.5644685 

1.73 

.9089692 

.7316847 

1.24 

.8724988 

.5831146 

1.74 

.9364319 

.7609115 

1.25 

.8884239 

.6019191 

1.75 

.9641884 

.7904143 

1.26 

.9045378 

.6208837 

1.76 

2.9922411 

.8201962 

1.27 

.9208421 

.6400105 

1.77 

3.0205932 

.8502601 

1.28 

.9373385 

.6593012 

1.78 

.0492473 

.8806091 

1.29 

.9540287 

.6787578  ' 

1.79 

.0782063 

.9112461 

1.30 

.9709143 

.6983824 

1.80 

.1074732 

.9421742 

1.31 

1.9879969 

.7181768 

1.81 

.1370508 

2.9733966 

1.32 

2.0052783 

.7381431 

1.82 

.1669421 

3.0049163 

1.33 

.0227603 

.7582830 

1.83 

.1971501 

.0367365 

1.34 

.0404446 

.7785989 

1.84 

.2276799 

.0688603 

1.35 

.0583329 

.7990926 

1.85 

.2585283 

.1012911 

1.36 

.0764271 

.8197662 

1.86 

.2897047 

.1340321 

1.37 

.0947288 

.8406219 

1.87 

.3212100 

.1670863 

1.38 

.1132401 

.8616615 

1.88 

.3530475 

.2004573 

1.39 

.1319627 

.8828874 

1.89 

.3852202 

.2341484 

1.40 

.1508985 

.9043015 

1.90 

.4177315 

.2681629 

1.41 

.1700494 

.9259060 

1.91 

.4505846 

.3025041 

1.42 

.1894172 

.9477032 

1.92 

.4837827 

.3371758 

1.43 

.2090041 

.9696951 

1.93 

.5173293 

.3721810 

1.44 

.2288118 

1.9918840 

1.94 

.5512275 

.4075235 

1.45 

.2488424 

2.0142721 

1.95 

.5854808 

.4432067 

1.46 

.2690979 

.0368616 

1.96 

.6200927 

.4792343 

1.47 

.2895803 

.0596549 

1.97 

.6550667 

.5156097 

1.48 

.3102917 

.082(5540 

1.98 

.6904061 

.5523368 

1.49 

.3312341 

.1058614 

1.99 

.7261146 

.5894191 

1.50 

2.3524096 

2.1292794 

2.00 

3.7621957 

3.6268604 

HYPERBOLIC  FUNCTIONS 


419 


X 

Cosh  a: 

Sinhz 

X 

Cosh  a; 

Sinha; 

2.01 

3.7986528 

3.6646642 

2.51 

6.1930993 

6.1118311 

2.02 

.8354899 

.7028345 

2.52 

.2545281 

.1740685 

2.03 

.8727101 

.7413746 

2.53 

.3165827 

.2369237 

2.04 

.9103184 

.7802896 

2.54 

.379268.7 

.3004023 

2.05 

.9483548 

.8196198 

2.55 

.4425928 

.3645111 

2.06 

.9867111 

.8592571 

2.56 

.5065611 

.4292563 

2.07 

4.0255038 

.8993179 

2.57 

.5711800 

.4946444 

2.08 

.0647395 

.9398093 

2.58 

.6364560 

.5606820 

2.09 

.1043012 

.9806140 

2.59 

.7023958 

.6273758 

2.10 

.1443131 

4.0218567 

2.60 

.7690059 

.6947323 

2.11 

.1847398 

.0635018 

2.61 

.8362940 

.7627595 

2.12 

.2255846 

.1055530 

2.62 

.9042644 

.8314615 

2.13 

.2668523 

.1480149 

2.63 

.9729254 

.9008469 

2.14 

.3085462 

.1908914 

2.64 

7.0422838 

.9709225 

2.15 

.3506713 

.2341871 

2.65 

.1123463 

7.0416950 

2.16 

.3932312 

.2779062 

2.66 

.1831184 

.1131701 

2.17 

.4362311 

.3220534 

2.67 

.2546108 

.1853586 

2.18 

.4796741 

.3666325 

2.68 

.3268282 

.2582650 

2.19 

.5235649 

.4116482 

2.69 

.3997785 

.3318975 

2.20 

.5679083 

.4571052 

2.70 

.4734686 

.4062631 

2.21 

.6127086 

.5030079 

2.71 

.5479060 

.4813692 

2.22 

.6579702 

.5493610 

2.72 

.6230984 

.5572237 

2.23 

.7036972 

.5961688 

2.73 

.6990531 

.6338338 

2.24 

.7498951 

.6434364 

2.74 

.7757775 

.7112072 

2.25 

.7965677 

.6911685 

2.75 

.8532799 

.7893520 

2.26 

.8437197 

.7393692 

2.76 

.9315674 

.8682756 

2.27 

.8913565 

.7880444 

2.77 

8.0106482 

.9479862 

2.28 

.9394824 

.8371982 

2.78 

.0905297 

8.0284911 

2.29 

.9881022 

.8868358 

2.79 

.1712205 

.1097993 

2.30 

5.0372206 

.9369618 

2.80 

.2527285 

.1919185 

2.31 

.0868429 

.9875817 

2.81 

.3350617 

.2748566 

2.32 

.1369741 

5.0387004 

2.82 

.4182283 

.3586224 

2.33 

.1876186 

.0903228 

2.83 

.5022368 

.4432239 

2.34 

.2387822 

.1424545 

2.84 

.5870956 

.5286699 

2.a5 

.2905196 

.1951504 

2.85 

.6728130 

.6149687 

2.3H 

.3426859 

.2482656 

2.86 

.7593979 

.7021291 

2.37 

.3954365 

.3019558 

2.87 

.8468585 

.7901595 

2.38 

.4487266 

.3561760 

2.88 

.9352041 

.8790694 

2.39 

.5025618 

.4109321 

2.89 

9.0244430 

.9688668 

2.40 

.5569472 

.4662293 

2.90 

.1145844 

9.0595611 

2.41 

.6118883 

.5220729 

2.91 

.2056373 

.1511616 

2.42 

.6673910 

.5784683 

2.92 

.2976105 

.2436769 

2.43 

.7234594 

.6354226 

2.93 

.3905138 

.3371168 

2.44 

.7801009 

.6929401 

2.94 

.4843559 

.4314902 

2.45 

.8373201 

.7510265 

2.95 

.5791467 

.5268070 

2.46 

.8951232 

.8096882 

2.96 

.6748952 

6230763 

2.47 

.9535159 

.8689310 

2.97 

.7716115 

.7203081 

2.48 

6.0125038 

.9287605 

2.98 

.8693047 

.8185119 

2.49 

.0720930 

.9891831 

2.99 

.9679850 

.9176976 

2.50 

6.1322895 

6.0502045 

3.00 

10.0676620 

10.0178750 

420 


HYPERBOLIC  FUNCTIONS 


X 

Coshx 

Sinha; 

X 

Cosh  a; 

Sinha; 

3.01 

10.1683456 

10.1190539 

3.51 

16.7390823 

16.7091854 

3.02 

.2700464 

.2212451 

3.52 

.9070139 

.8774144 

3.03 

.3727741 

.3244585 

3.53 

17.0766361 

17.0473312 

3.04 

.4765391 

.4287042 

3.54 

.2479662 

.2189529 

3.05 

.5813518 

.5339929 

3.55 

.4210213 

.3922966 

3.06 

.6872224 

.6403347 

3.56 

.5958178 

.5673790 

3.07 

.7941620 

.7477408 

3.57 

.7723744 

.7442186 

3.08 

.9021809 

.8562217 

3.58 

.9507082 

.9228325 

3.09 

11.0112900 

.9657881 

3.59 

18.1308371 

18.1032388 

3.10 

.1215004 

11.0764511 

3.60 

.3127790 

.2854552 

3.11 

.2328226 

.1882217 

3.61  - 

.4965523 

.4695004 

3.12 

.3452684 

.3011112 

3.62 

.6821753 

.6553927 

3.13 

.4588488 

.4151309 

3.63 

.8696665 

.8431503 

3.14 

.5735748 

.5302919 

3.64 

19.0590447 

19.0327924 

3.15 

.6894584 

.6466062 

3.65 

.2503288 

.2243376 

3.16 

.8065107 

.7640850 

3.66 

.4435377 

.4178052 

3.17 

.9247440 

.8827403 

3.67 

.6386909 

.6132145 

3.18 

12.0441695 

12.0025838 

3.68 

.8358083 

.8105854 

3.19 

.1647998 

.1236279 

3.69 

20.0349094 

20.0099373 

3.20 

.2866462 

.2458839 

3.70 

.2360140 

.2112905 

3.21 

.4097213 

.3693646 

3.71 

.4391421 

.4146645 

3.22 

.5340375 

.4940825 

3.72 

.6443142 

.6200802 

3.23 

.6596073 

.6200497 

3.73 

.8515505 

.8275577 

3.24 

.7864428 

.7472790 

3.74 

21.0608720 

21.0371178 

3.25 

.9145572 

.8757829 

3.75 

.2722997 

.2487819 

3.26 

13.0439629 

13.0055744 

3.76 

.4858548 

.4625710 

3.27 

.1746730 

.136(5665 

3.77 

.7015584 

.6785064 

3.28 

.3067006 

.2690723 

3.78 

.9194324 

.8966096 

3.29 

.4400587 

.4028048 

3.79 

22.1394981 

22.1169025 

3.30 

.5747611 

.5378780 

3.80 

.3617777 

.3394069 

3.31 

.7108208 

.6743046 

3.81 

.5862933 

.5641452 

3.32 

.8482516 

.8120988 

3.82 

.8130681 

.7911403 

3.33 

.9870673 

.9512741 

3.83 

23.0421239 

23.0204143 

3.34 

14.1272820 

14.0918450 

3.84 

.2734843 

.2519907 

3.35 

.2689091 

.2338247 

3.85 

.5071715 

.4858917 

3.36 

.4119630 

.3772277 

3.86 

.7432095 

.7221415 

3.37 

.5564583 

.5220686 

3.87 

.9816222 

.9607638 

3.38 

.7024094 

.6683619 

3.88 

24.2224327 

24.2017819 

3.39 

.8498306 

.8161219 

3.89 

.4656658 

.4452205 

3.40 

.9987366 

.9653634 

3.90 

.7113454 

.6911034 

3.41 

15.1491429 

15.1161016 

3.91 

.9594963 

.9394557 

3.42 

.3010637 

.2683513 

3.92 

25.2101431 

25.1903020 

3.43 

.4545147 

.4221278 

3.93 

.4633109 

.4436673 

3.44 

.6095114 

.5774468 

3.94 

.7190247 

.6995765 

3.45 

.7660688 

.7343232 

3.95 

.9773109 

.9580561 

3.46 

.9242033 

.8927735 

3.96 

26.2381943 

26.2191311 

3.47 

16.0839298 

16.0528128 

3.97 

.5017019 

.4828285 

3.48 

.2452646 

.2144571 

3.98 

.7678597 

.7491740 

3.49 

.4082241 

.3777233 

3.99 

27.0366943 

27.0181946 

3.50 

16.5728248 

16.5426275 

4.00 

.3082331 

.2899175 

APPENDIX  II 
LOGARITHMS   OF   NUMBERS 


422 


LOGARITHMS   OF  NUMBERS 


LOGARITHMS  OF  NUMBERS,  PROM  O  TO  1OOO 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0 

-co 

00000 

30103 

47712 

60206 

69897 

77815 

84510 

90309 

95424 

10 

00000 

00432 

00860 

01283 

01703 

02118 

02530 

02938 

03342 

03742 

11 

04139 

04532 

04921 

05307 

05690 

06069 

06445 

06818 

07188 

07554 

12 

07918 

08278 

08636 

08990 

09342 

09691 

10037 

10380 

10721 

11059 

13 

11394 

11727 

12057 

12385 

12710 

13033 

13353 

13672 

13987 

14301 

14 

14613 

14921 

15228 

15533 

15836 

16136 

16435 

16731 

17026 

17318 

15 

17609 

17897 

18184 

18469 

18752 

19033 

19312 

19590 

19865 

20139 

16 

20412 

20682 

20951 

21218 

21484 

21748 

22010 

22271 

22530 

22788 

17 

23045 

23299 

23552 

23804 

24054 

24303 

24551 

24797 

25042 

25285 

18 

25527 

25767 

26007 

26245 

26481 

26717 

26951 

27184 

27415 

27646 

19 

27875 

28103 

28330 

28555 

28780 

29003 

29225 

29446 

29666 

29885 

20 

30103 

30319 

30535 

30749 

30963 

31175 

31386 

31597 

31806 

32014 

21 

32222 

32428 

32633 

32838 

33041 

33243 

33445 

33646 

33845 

34044 

22 

34242 

34439 

34635 

34830 

35024 

35218 

35410 

35602 

35793 

35983 

23 

36173 

36361 

36548 

36735 

36921 

37106 

37291 

37474 

37657 

37839 

24 

38021 

38201 

38381 

38560 

38739 

38916 

39093 

39269 

39445 

39619 

25 

39794 

39967 

40140 

40312 

40483 

40654 

40824 

40993 

•41162 

41330 

26 

41497 

41664 

41830 

41995 

42160 

42324 

42488 

42651 

42813 

42975 

27 

43136 

43296 

43456 

43616 

43775 

43933 

44090 

44248 

44404 

44560 

28 

44716 

44870 

45024 

45178 

45331 

45484 

45636 

45788 

45939 

46089 

29 

46240 

46389 

46538 

46686 

46834 

46982 

47129 

47275 

47421 

47567 

30 

47712 

47856 

48000 

48144 

48287 

48430 

48572 

48713 

48855 

48995 

31 

49136 

49276 

49415 

49554 

49693 

49831 

49968 

50105 

50242 

50379 

32 

50515 

50650 

50785 

50920 

51054 

51188 

51321 

51454 

51587 

51719 

33 

51851 

51982 

52113 

52244 

52374 

52504 

52633 

52763 

52891 

53020 

34 

53148 

53275 

53402 

53529 

53655 

53781 

53907 

54033 

54157 

54282 

35 

54407 

54530 

54654 

54777 

54900 

55022 

55145 

55266 

55388 

55509 

36 

55630 

55750 

55870 

55990 

56110 

56229 

56348 

56466 

56584 

56702 

37 

56820 

56937 

57054 

57170 

57287 

57403 

57518 

57634 

57749 

57863 

38 

57978 

58092 

58206 

58319 

58433 

58546 

58658 

58771 

58883 

58995 

39 

59106 

59217 

59328 

59439 

59549 

59659 

59769 

59879 

59988 

60097 

40 

60206 

60314 

60422 

60530 

60638 

60745 

60852 

60959 

61066 

61172 

41 

61278 

61384 

61489 

61595 

61700 

61804 

61909 

62013 

62118 

62221 

42 

62325 

62428 

62531 

62634 

62736 

62838 

62941 

63042 

63144 

63245 

43 

63347 

63447 

63548 

63648 

63749 

63848 

63948 

64048 

64147 

64246 

44 

64345 

64443 

64542 

64640 

64738 

64836 

64933 

65030 

65127 

65224 

45 

65321 

65417 

65513 

65609 

65705 

65801 

65896 

65991 

66086 

66181 

46 

66276 

66370 

66464 

66558 

66651 

66745 

66838 

66931 

67024 

67117 

47 

67210 

67302 

67394 

67486 

67577 

67669 

67760 

67851 

67942 

68033 

48 

68124 

68214 

68304 

68394 

68484 

68574 

68663 

68752 

68842 

68930 

49 

69020 

69108 

69196 

69284 

69372 

69460 

69548 

69635 

69722 

69810 

50 

69897 

69983 

70070 

70156 

70243 

70329 

70415 

70500 

70586 

70671 

51 

70757 

70842 

70927 

71011 

71096 

71180 

71265 

71349 

71433 

71516 

52 

71600 

71683 

71767 

71850 

71933 

72015 

72098 

72181 

72263 

72345 

53 

72428 

72509 

72591 

72672 

72754 

72835 

72916 

72997 

73078 

73158 

54 

73239 

73319 

73399 

73480 

73559 

73639 

73719 

73798 

73878 

73957 

LOGARITHMS   OF  NUMBERS 


423 


LOGARITHMS  OP  NUMBERS,  PROM  O  TO  1OOO 

(Continued) 

No. 

0 

1 

2 

3 

4 

5 

6 

7    8 

9 

55 

74036 

74115 

74193 

74272 

74351 

74429 

74507 

74585 

74663 

74741 

56 

74818 

74896 

74973 

75050 

75127 

75204 

75281 

75358 

75434 

75511 

57 

75587 

75663 

75739 

75815 

75891 

75966 

76042 

76117 

76192 

76267 

58 

76342 

76417 

76492 

76566 

76641 

76715 

76789 

76863 

76937 

77011 

59 

77085 

77158 

77232 

77305 

77378 

77451 

77524 

77597 

77670 

77742 

60 

77815 

77887 

77959 

78031 

78103 

78175 

78247 

78318 

78390 

78461 

61 

78533 

78604 

78675 

78746 

78816 

78887 

78958 

79028 

79098 

79169 

62 

79239 

79309 

79379 

79448 

79518 

79588 

79657 

79726 

79796 

79865 

63 

79934 

80002 

80071 

80140 

80208 

80277 

80345 

80413 

80482 

80550 

64 

80618 

80685 

80753 

80821 

80888 

80956 

81023 

81090 

81157 

81224 

65 

81291 

81358 

81424 

81491 

81557 

81624 

81690 

81756 

81822 

81888 

66 

81954 

82020 

82085 

82151 

82216 

82282 

82347 

82412 

82477 

82542 

67 

82607 

82672 

82736 

82801 

82866 

82930 

82994 

83058 

83123 

83187 

68 

83250 

83314 

83378 

83442 

83505 

83569 

83632 

83695 

83758 

83821 

69 

83884 

83947 

84010 

84073 

84136 

84198 

84260 

84323 

84385 

84447 

70 

84509 

84571 

84633 

84695 

84757 

84818 

84880 

84941 

85003 

85064 

71 

85125 

85187 

85248 

85309 

85369 

85430 

85491 

85551 

85612 

85672 

72 

85733 

85793 

85853 

85913 

85973 

86033 

86093 

86153 

86213 

86272 

73 

86332 

86391 

86451 

86510 

86569 

86628 

86687 

86746 

86805 

86864 

74 

86923 

86981 

87040 

87098 

87157 

87215 

87273 

87332 

87390 

87448 

75 

87506 

87564 

87621 

87679 

87737 

87794 

87852 

87909 

87966 

88024 

76 

88081 

88138 

88195 

88252 

88309 

88366 

88422 

88479 

88536 

88592 

77 

88649 

88705 

88761 

88818 

88874 

88930 

88986 

89042 

89098 

89153 

78 

89209 

89265 

89320 

89376 

89431 

89487 

89542 

89597 

89652 

89707 

79 

89762 

89817 

89872 

89927 

89982 

90036 

90091 

90145 

90200 

90254 

80 

90309 

90363 

90417' 

90471 

90525 

90579 

90633 

90687 

90741 

90794 

81 

90848 

90902 

90955 

91009 

91062 

91115 

91169 

91222 

91275 

91328 

82 

91381 

91434 

91487 

91540 

91592 

91645 

91698 

91750 

91803 

91855 

83 

91907 

91960 

92012 

92064 

92116 

92168 

92220 

92272 

92324 

92376 

84 

92427 

92479 

92531 

92582 

92634 

92685 

92737 

92788 

92839 

92890 

85 

92941 

92993 

93044 

93095 

93146 

93196 

93247 

93298 

93348 

93399 

86 

93449 

93500 

93550 

93601 

93651 

93701 

93751 

93802 

93852 

93902 

87 

93951 

94001 

94051 

94101 

94151 

94200 

94250 

94300 

94349 

94398 

88 

94448 

94497 

94546 

94596 

94645 

94694 

94743 

94792 

94841 

94890 

89 

94939 

94987 

95036 

95085 

95133 

95182 

95230 

95279 

95327 

95376 

90 

95424 

95472 

95520 

95568 

95616 

95664 

95712 

95760 

95808 

95856 

91 

95904 

95951 

95999 

96047 

96094 

96142 

96189 

96236 

96284 

96331 

92 

96378 

96426 

96473 

96520 

96567 

96614 

96661 

96708 

96754 

96801 

93 

96848 

96895 

96941 

96988 

97034 

97081 

97127 

97174 

97220 

97266 

94 

97312 

97359 

97405 

97451 

97497 

97543 

97589 

97635 

97680 

97726 

95 

97772 

97818 

97863 

97909 

97954 

98000 

98045 

98091 

98136 

98181 

96 

98227 

98272 

98317 

98362 

98407 

98452 

98497 

98542 

98587 

98632 

97 

98677 

98721 

98766 

98811 

98855 

98900 

98945 

98989 

99033 

99078 

98 

99122 

99166 

99211 

99255 

99299 

99343 

99387 

99431 

99475 

99519 

99 

99563 

99607 

99651 

99694 

99738 

99782 

99825 

99869 

99913 

99956 

APPENDIX  III 
TRIGONOMETRIC  FUNCTIONS 


TRIGONOMETRIC  FUNCTIONS 
1 


427 


NATURAL  SINES,  COSINES,   TANGENTS,  ETC. 

o 

/ 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

0 

0 

0 

.000000 

Infinite 

.000000 

Infinite 

1.00000 

1.000000 

0 

90 

10 

.002909 

343.77516 

.002909 

343.77371 

1.00000 

.999996 

50 

20 

.005818 

171.88831 

.005818 

171.88540 

1.00002 

.999983 

40 

30 

.008727 

114.59301 

.008727 

114.58865 

1.00004 

.999962 

30 

40 

.011635 

85.945609 

.011636 

85.939791 

1.00007 

.999932 

20 

50 

.014544 

68.757360 

.014545 

68.750087 

1.00011 

.999894 

10 

1 

0 

.017452 

57.298688 

.017455 

57.289962 

1.00015 

.999848 

0 

89 

10 

.020361 

49.114062 

.020365 

49.103881 

1.00021 

.999793 

50 

20 

.023269 

42.975713 

.023275 

42.964077 

1.00027 

.999729 

40 

30 

.026177 

38.201550 

.026186 

38.188459 

1.00034 

.999657 

30 

40 

.029085 

34.382316 

.0129097 

34.367771 

1.00042 

.999577 

20 

50 

.031992 

31.257577 

.032009 

31.241577 

1.00051 

.999488 

10 

2 

0 

.034899 

28.653708 

.034921 

28.636253 

1.00061 

.999391 

0 

88 

10 

.037806 

26.450510 

.037834 

26.431600 

1.00072 

.999285 

50 

20 

.040713 

24.562123 

.040747 

24.541758 

1.00083 

.999171 

40 

30 

.043619 

22.925586 

.043661 

22.903766 

1.00095 

.999048 

30 

40 

.046525 

21.493676 

.046576 

21.470401 

1.00108 

.998917 

20 

50 

.049431 

20.230284 

.049491 

20.205553 

1.00122 

.998778 

10 

3 

0 

.052336 

19.107323 

.052408 

19.081137 

1.00137 

.998630 

0 

87 

10 

.055241 

18.102619 

.055325 

18.074977 

1.00153 

.998473 

50 

20 

.058145 

17.1984:34 

.058243 

17.169337 

1.00169 

.998308 

40 

30 

.061049 

16.380408 

.061163 

16.349855 

1.00187 

.998135 

30 

40 

.063952 

15.636793 

.064083 

15.604784 

1.00205 

.997357 

20 

50 

.066854 

14.957882 

.0(37004 

14.924417 

1.00224 

.997763 

10 

4 

0 

.069756 

14.335587 

.069927 

14.300666 

1.00244 

.997564 

0 

86 

10 

.072658 

13.763115 

.072851 

13.726738 

1.00265 

.997357 

50 

20 

.075559 

13.234717 

.075776 

13.196888 

1.00287 

.997141 

40 

30 

.078459 

12.745495 

.078702 

12.706205 

1.00309 

.996917 

30 

40 

.081359 

12.291252 

.081629 

12.250505 

1.00333 

.996685 

20 

50 

.084258 

11.868370 

.084558 

11.826167 

1.00357 

.996444 

10 

5 

0 

.087156 

11.473713 

.087489 

11.430052 

1.00382 

.996195 

0 

85 

10 

.090053 

11.104549 

.090421 

11.059431 

1.00408 

.995937 

50 

20 

.092950 

10.758488 

.093354 

10.711913 

1.00435 

.995671 

40 

30 

.095846 

10.433431 

.096289 

10.385397 

1.00463 

.995396 

30 

40 

.098741 

10.127522 

.099226 

10.078031 

1.00491 

.995113 

20 

50 

.101635 

9.8391227 

.102164 

9.7881732 

1.00521 

.994822 

10 

6 

0 

.104528 

9.5667722 

.105104 

9.5143645 

1.00551 

.994522 

0 

84 

10 

.107421 

9.3091699 

.108046 

9.2553035 

1.00582 

.994214 

50 

20 

.110313 

9.0651512 

.110990 

9.0098261 

1.00614 

.993897 

40 

83 

o 

f 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

I 

o 

For  functions  from  83°  40'  to  90°  read  from  bottom  of  table  upward. 

428 


TRIGONOMETRIC  FUNCTIONS 


NATURAL  SINES,  COSINES,  TANGENTS,  ETC. 

j        (Continued} 

0 

1 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

0 

6 

30 

.113203 

8.8336715 

.113936 

8.7768874 

1.00647 

.993572 

30 

40 

.116093 

8.6137901 

.116883 

8.5555468 

1.00681 

.993238 

20 

50 

.118982 

8.4045586 

.119833 

8.3449558 

1.00715 

.992896 

10 

7 

0 

.121869 

8.2055090 

.122785 

8.1443464 

1.00751 

.992546 

0 

83 

10 

.124756 

8.0156450 

.125738 

7.9530224 

1.00787 

.992187 

50 

20 

.127642 

7.8344335 

.128694 

7.7703506 

1.00825 

.991820 

40 

30 

.130526 

7.6612976 

.131653 

7.5957541 

1.00863 

.991445 

30 

40 

.133410 

7.4957100 

.134613 

7.4287064 

1.00902 

.991061 

20 

50 

.136292 

7.3371909 

.137576 

7.2687255 

1.00942 

.990669 

10 

8 

0 

.139173 

7.1852965 

.140541 

7.1153697 

1.00983 

.990268 

0 

82 

10 

.142053 

7.0396220 

.143508 

6.9682335 

1.01024 

.989859 

50 

20 

.144932 

6.8997942 

.146478 

6.8269437 

1.01067 

.989442 

40 

30 

.147809 

6.7654691 

.149451 

6.6911562 

1.01111 

.989016 

30 

40 

.150686 

6.6363293 

.152426 

6.5605538 

1.01155 

.988582 

20 

50 

.153561 

6.5120812 

.155404 

6.4348428 

1.01200 

.988139 

10 

9 

0 

.156434 

6.3924532 

.158384 

6.3137515 

1.01247 

.987688 

0 

81 

10 

.159307 

6.2771933 

.161368 

6.1970279 

1.01294 

.987229 

50 

20 

.162178 

6.1660674 

.164354 

6.0844381 

1.01342 

.986762 

40 

30 

.165048 

6.0588980 

.167343 

5.9757644 

1.01391 

.986286 

30 

40 

.167916 

5.9553625 

.170334 

5.8708042 

1.01440 

.985801 

20 

50 

.170783 

5.8553921 

.173329 

5.7693688 

1.01491 

.985309 

10 

10 

0 

.173648 

5.7587705 

.176327 

5.6712818 

1.01543 

.984808 

0 

80 

10 

.176512 

5.6653331 

.179328 

5.5763786 

1.01595 

.984298 

50 

20 

.179375 

5.5749258 

.182332 

5.4845052 

1.01649 

.983781 

40 

30 

.182236 

5.4874043 

.185339 

5.3955172 

1.01703 

.983255 

30 

40 

.185095 

5.4026333 

.188359 

5.3092793 

1.01758 

.982721 

20 

50 

.187953 

5.3204860 

.191363 

5.2256647 

1.01815 

.982178 

10 

11 

0 

.190809 

5.2408431 

.194380 

5.1445540 

1.01872 

.981627 

0 

79 

10 

.193664 

5.1635924 

.197401 

5.0658352 

1.01930 

.981068 

50 

20 

.196517 

5.0886284 

.200425 

4.9894027 

1.01989 

.980500 

40 

30 

.199368 

5.0158317 

.203452 

4.9151570 

1.02049 

.979925 

30 

40 

.202218 

4.9451687 

.206483 

4.8430045 

1.02110 

.979341 

20 

50 

.205065 

4.8764907 

.209518 

4.7728568 

1.02171 

.978748 

10 

12 

0 

.207912 

4.8097343 

.212557 

4.7046301 

1.02234 

.978148 

0 

78 

10 

.210756 

4.7448206 

.215599 

4.6382457 

1.02298 

.977539 

50 

20 

.213599 

4.6816748 

.218645 

4.5736287 

1.02362 

.976921 

40 

30 

.216440 

4.6202263 

.221695 

4.5107085 

1.02428 

.976296 

30 

40 

.219279 

4.5604080 

.224748 

4.4494181 

1.02494 

.975662 

20 

50 

.222116 

4.5021565 

.227806 

4.3896940 

1.02562 

.975020 

10 

77 

o 

/ 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

I 

o 

For  functions  from  77°  10'  to  83°  30'  read  from  bottom  of  table  upward. 

TRIGONOMETRIC  FUNCTIONS 


429 


NATURAL  SINES,  COSINES,   TANGENTS,  ETC. 

(Continued) 

o 

t 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

0 

13 

0 

.224951 

4.4454115 

.230868 

4.3314759 

1.02630 

.974370 

0 

77 

10 

.227784 

4.3901158 

.233934 

4.2747066 

1.02700 

.973712 

50 

20 

.230616 

4.3362150 

.237004 

4.2193318 

1.02770 

.973045 

40 

30 

.233445 

4.2836576 

.240079 

4.1652998 

1.02842 

.972370 

30 

40 

.236273 

4.2323943 

.243158 

4.1125614 

1.02914 

.971687 

20 

50 

.239098 

4.1823785 

.246241 

4.0610700 

1.02987 

.970995 

10 

14 

0 

.241922 

4.1335655 

.249328 

4.0107809 

1.03061 

.970296 

0 

76 

10 

.244743 

4.0859130 

.252420 

3.9616518 

1.03137 

.969588 

50 

20 

.247563 

4.0393804 

.255517 

3.9136420 

1.03213 

.968872 

40 

30 

.250380 

3.9939292 

.258618 

3.8667131 

1.03290 

.968148 

30 

40 

.253195 

3.9495224 

.261723 

3.8208281 

1.03363 

.967415 

20 

50 

.256008 

3.9061250 

.264834 

3.7759519 

1.03447 

.966675 

10 

15 

0 

.258819 

3.8637033 

.267949 

3.7320508 

1.03528 

.965926 

0 

75 

10 

.261628 

3.8222251 

.271069 

3.6890927 

1.03609 

.965169 

50 

20 

.264434 

3.7816596 

.274195 

3.6470467 

1.03691 

.964404 

40 

30 

.267238 

3.7419775 

.277325 

3.6058835 

1.03774 

.963630 

30 

40 

.270040 

3.7031506 

.280460 

3.5655749 

1.03858 

.962849 

20 

50 

.272840 

3.6651518 

.283600 

3.5260938 

1.03944 

.962059 

10 

16 

0 

.275637 

3.6279553 

.286745 

3.4874144 

1.04030 

.961262 

0 

74 

10 

.278432 

3.5915363 

.289896 

3.4495120 

1.04117 

.960456 

50 

20 

.281225 

3.5558710 

.293052 

3.4123626 

1.04206 

.959642 

40 

30 

.284015 

3.5209365 

.296214 

3.3759434 

1.04295 

.958820 

30 

40 

.286803 

3.4867110 

.299380 

3.3402326 

1.04385 

.957990 

20 

50 

.289589 

3.4531735 

.302553 

3.3052091 

1.04477 

.957151 

10 

17 

0 

.292372 

3.4203036 

.305731 

3.2708526 

1.04569 

.956305 

0 

73 

10 

.295152 

3.3880820 

.308914 

3.2371438 

1.04663 

.955450 

50 

20 

.297930 

3.3564900 

.312104 

3.2040638 

1.04757 

.954588 

40 

30 

.300706 

3.3255095 

.315299 

3.1715948 

1.04853 

.953717 

30 

40 

.303479 

3.2951234 

.318500 

3.1397194 

1.04950 

.952838 

20 

50 

.306249 

3.2653149 

.321707 

3.1084210 

1.05047 

.951951 

10 

18 

0 

.309017 

3.2360680 

.324920 

3.0776835 

1.05146 

.951057 

0 

72 

10 

.311782 

3.2073673 

.328139 

3.0474915 

1.05246 

.950154 

50 

20 

.314545 

3.1791978 

.331364 

3.0178301 

1.05347 

.949243 

40 

30 

.317305 

3.1515453 

.334595 

2.9886850 

1.05449 

.948324 

30 

40 

.320062 

3.1243959 

.337833 

2.9600422 

1.05552 

.947397 

20 

50 

.322816 

3.0977363 

.341077 

2.9318885 

1.05657 

.946462 

10 

19 

0 

.325568 

3.0715535 

.344328 

2.9042109 

1.05762 

.945519 

0 

71 

10 

.328317 

3.0458352 

.347585 

2.8769970 

1.05869 

.944568 

50 

20 

.331063 

3.0205693 

.350848 

2.8502349 

1.05976 

.943609 

40 

70 

0 

/ 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

I 

0 

For  functions  from  70°  40'  to  77°  0'  read  from  bottom  of  table  upward. 

430 


TRIGONOMETRIC  FUNCTIONS 


NATURAL  SINES,  COSINES,  TANGENTS,  ETC. 

(  Continued) 

o 

l 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

t 

o 

19 

30 

.333807 

2.9957443 

.354119 

2.8239129 

1.06085 

.942641 

30 

40 

.336547 

2.9713490 

.357396 

2.7980198 

1.06195 

.941666 

20 

50 

.339285 

2.9473724 

.360680 

2.7725448 

1.06306 

.940684 

10 

20 

0 

.342020 

2.9238044 

.363970 

2.7474774 

1.06418 

.939693 

0 

70 

10 

.344752 

2.9006346 

.367268 

2.7228076 

1.06531 

.938694 

50 

20 

.347481 

2.8778532 

.370573 

2.6985254 

1.06645 

.937687 

40 

30 

.350207 

2.8554510 

.373885 

2.6746215 

1.06761 

.936672 

30 

40 

.352931 

2.8334185 

.377204 

2.6510867 

1.06878 

.935650 

20 

50 

.355651 

2.8117471 

.380530 

2.6279121 

1.06995 

.934619 

10 

21 

0 

.358368 

2.7904281 

.383864 

2.6050891 

1.07115 

.93&580 

0 

69 

10 

.361082 

2.7694532 

.387205 

2.5826094 

1.07235 

.932534 

50 

20 

.363793 

2.7488144 

.390554 

2.5604649 

1.07356 

.931480 

40 

30 

.366501 

2.7285038 

.393911 

2.5386479 

1.07479 

.930418 

30 

40 

.369206 

2.7085139 

.397275 

2.5171507 

1.07602 

.929348 

20 

50 

.371908 

2.6888374 

.400647 

2.4959661 

1.07727 

.928270 

10 

22 

0 

.374607 

2.6694672 

.404026 

2.4750869 

1.07853 

.927184 

0 

68 

10 

.377302 

2.6503962 

.407414 

2.4545061 

1.07981 

.926090 

50 

20 

.379994 

2.6316180 

.410810 

2.4342172 

1.08109 

.924989 

40 

30 

.382683 

2.6131259 

.414214 

2.4142136 

1.08239 

.923880 

30 

40 

.385369 

2.5949137 

.417626 

2.3944889 

1.08370 

.922762 

20 

50 

.388052 

2.5769753 

.421046 

2.3750372 

1.08503 

.921638 

10 

23 

0 

.390731 

2.5593047 

.424475 

2.3558524 

1.08636 

.920505 

0 

67 

10 

.393407 

2.5418961 

.427912 

2.3369287 

1.08771 

.919364 

50 

20 

.396080 

2.5247440 

.431358 

2.3182606 

1.08907 

.918216 

40 

30 

.398749 

2.5078428 

.434812 

2.2998425 

1.09044 

.917060 

30 

40 

.401415 

2.4911874 

.438276 

2.2816693 

1.09183 

.915896 

20 

50 

.404078 

2.4747726 

.441748 

2.2637357 

1.09323 

.914725 

10 

24 

0 

.406737 

2.4585933 

.445229 

2.2460368 

1.09464 

.913545 

0 

66 

10 

.409392 

2.4426448 

.448719 

2.2285676 

1.09606 

.912358 

50 

20 

.412045 

2.4269222 

.452218 

2.2113234 

1.09750 

.911164 

40 

30 

.414693 

2.4114210 

.455726 

2.1942997 

1.09895 

.909961 

30 

40 

.417338 

2.3961367 

.459244 

2.1774920 

1.10041 

.908751 

20 

50 

.419980 

2.3810650 

.462771 

2.1608958 

1.10189 

.907533 

10 

25 

0 

.422618 

2.3662016 

.466308 

2.1445069 

1.10338 

.906308 

0 

65 

10 

.425253 

2.3515424 

.469854 

2.1283213 

1.10488 

.905075 

50 

20 

.427884 

2.3370833 

.473410 

2.1123348 

1.10640 

.903834 

40 

30 

.430511 

2.3228205 

.476976 

2.0965436 

1.10793 

.902585 

30 

40 

.433135 

2.3087501 

.480551 

2.0809438 

1.10947 

.901329 

20 

50 

.435755 

2.2948685 

.484137 

2.0655318 

1.11103 

.900065 

10 

64 

o 

? 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

t 

o 

For  functions  from  64°-10'  to  70°-30'  read  from  bottom  of  table  upward. 

TRIGONOMETRIC  FUNCTIONS 


431 


NATURAL  SINES,   COSINES,  TANGENTS,  ETC. 

(Continued} 

o 

/ 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

o 

26 

0 

.438371 

2.2811720 

.487733 

2.0503038 

1.11260 

.898794 

0 

64 

10 

.440984 

2.2676571 

.491339 

2.0352565 

1.11419 

.897515 

50 

20 

.443593 

2.2543204 

.494955 

2.0203862 

1.11579 

.896229 

40 

30 

.446198 

2.2411585 

.498582 

2.0056897 

1.11740 

.894934 

30 

40 

.448799 

2.2281681 

.502219 

1.9911637 

1.11903 

.893633 

20 

50 

.451397 

2.2153460 

.505867 

1.9768050 

1.12067 

.892323 

10 

27 

0 

.453990 

2.2026893 

.509525 

1.9626105 

1.12233 

.891007 

0 

63 

10 

.456580 

2.1901947 

.513195 

1.9485772 

1.12400 

.889682 

50 

20 

.459166 

2.1778595 

.516876 

1.9347020 

1.12568 

.888350 

40 

30 

.461749 

2.1656806 

.520567 

1.9209821 

1.12738 

.887011 

30 

40 

.464327 

2.1536553 

,524270 

1.9074147 

1.12910 

.885664 

20 

50 

.466901 

2.1417808 

.527984 

1.8939971 

1.13083 

.884309 

10 

28 

0 

.469472 

2.1300545 

.531709 

1.8807265 

1.13257 

.882948 

0 

62 

10 

.472038 

2.1184737 

.535547 

1.8676003 

1.13433 

.881578 

50 

20 

.474600 

2.1070359 

.539195 

1.8546159 

1.13610 

.880201 

40 

30 

.477159 

2.0957385 

.542956 

1.8417409 

1.13789 

.878817 

30 

40 

.479713 

2.0845792 

.546728 

1.8290628 

1.13970 

.877425 

20 

50 

.482263 

2.0735556 

.550515 

1.8164892 

1.14152 

.876026 

10 

29 

0 

.484810 

2.0626653 

.554309 

.8040478 

1.14335 

.874620 

0 

61 

10 

.487352 

2.0519061 

.558118 

.7917362 

1.14521 

.873206 

50 

20 

.489890 

2.0412757 

.561939 

.7795524 

1.14707 

.871784 

40 

30 

.492424 

2.0307720 

.565773 

.7674940 

.14896 

.870356 

30 

40 

.494953 

2.0203929 

.569619 

.7555590 

.15085 

.868920 

20 

50 

.497479 

2.0101362 

.573478 

.7437453 

.15277 

.867476 

10 

30 

0 

.500000 

2.0000000 

.577350 

1.7320508 

.15470 

.866025 

0 

60 

10 

.502517 

1.9899822 

.581235 

1.7204736 

.15665 

.864567 

50 

20 

.505030 

1.9800810 

.585134 

1.7090116 

.15861 

.863102 

40 

30 

.507538 

1.9702944 

.589045 

1.6976631 

1.16059 

.861629 

30 

40 

.510043 

1.9606206 

.592970 

1.6864261 

1.16259 

.860149 

20 

50 

.512543 

1.9510577 

.596908 

1.6752988 

1.16460 

.858662 

10 

31 

0 

.515038 

1.9416040 

.600861 

1.6642795 

1.16663 

.857167 

0 

59 

10 

.517529 

1.9322578 

.604827 

1.6533663 

1.16868 

.855665 

50 

20 

.520016 

1.9230173 

.608807 

1.6425576 

1.17075 

.854156 

40 

30 

.522499 

1.9138809 

.612801 

.6318517 

1.17283 

.852640 

30 

40 

.524977 

1.9048469 

.616809 

.6212469 

1.17493 

.851117 

20 

50 

.527450 

1.8959138 

.620832 

.6107417 

1.17704 

.849586 

10 

32 

0 

.529919 

1.8870799 

.624869 

1.6003345 

1.17918 

.848048 

0 

58 

10 

.532384 

1.8783438 

.628921 

.5900238 

1.18133 

.846503 

50 

20 

.534844 

1.8697040 

.632988 

1.5798079 

1.18350 

.844951 

40 

57 

0 

I 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

I 

o 

For  functions  from  57°-40'  to  64°-0'  read  from  bottom  of  table  upward. 

432 


TRIGONOMETBIC  FUNCTIONS 
6 


NATURAL  SINES,  COSINES,  TANGENTS,  ETC. 

(Continued) 

0 

/ 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

0 

32 

30 

.537300 

1.8611590 

'.637079 

1.5696856 

1.18569 

.843391 

30 

40 

.539751 

1.8527073 

.641167 

1.5596552 

1.18790 

.841825 

20 

50 

.542197 

1.8443476 

.645280 

1.5497155 

1.19012 

.840251 

10 

33 

0 

.544639 

1.8360785 

.649408 

1.5398650 

1.19236 

.838671 

0 

57 

10 

.547076 

1.8278985 

.653531 

1.5301025 

1.19463 

.837083 

50 

20 

.549509 

1.8198065 

.657710 

1.5204261 

1.19691 

.835488 

40 

30 

.551937 

1.8118010 

.661886 

1.5108352 

1.19920 

.833886 

30 

40 

.554360 

1.8038809 

.666077 

1.5013282 

1.20152 

.832277 

20 

50 

.556779 

1.7960449 

.670285 

1.4919039 

1.20386 

830661 

10 

34 

0 

.559193 

1.7882916 

.674509 

1.4825610 

1.20622 

.829038 

0 

56 

10 

.561602 

1.7806201 

.678749 

1.4732983 

1.20859 

.827407 

50 

20 

.564007 

1.7730290 

.683007 

1.4641147 

1.21099 

.825770 

40 

30 

.566406 

1.7655173 

.687281 

1.4550090 

1.21341 

.824126 

30 

40 

.568801 

1.7580837 

.691573 

1.4459801 

1.21584 

.822475 

20 

50 

.571191 

1.7507273 

.695881 

1.4370268 

1.21830 

.820817 

10 

35 

0 

.573576 

1.7434468 

.700208 

1.4281480 

1.22077 

.819152 

0 

55 

10 

.575957 

1.7362413 

.704552 

1.4193427 

1.22327 

.817480 

50 

20 

.578332 

1.7291096 

.708913 

1.4106098 

1.22579 

.815801 

40 

30 

.580703 

1.7220508 

.713293 

1.4019483 

1.22833 

.814116 

30 

40 

.583069 

1.7150639 

.717691 

1.3933571 

1.23089 

.812423 

20 

50 

.585429 

1.7081478 

.722108 

1.3848355 

1.23347 

.810723 

10 

36 

0 

.587785 

1.7013016 

.726543 

1.3763810 

1.23607 

.809017 

0 

54 

10 

.590136 

1.6945244 

.730996 

1.3679959 

1.23869 

.807304 

50 

20 

.592482 

1.6878151 

.735469 

1.  359(5764 

1.24134 

.805584 

40 

30 

.594823 

1.6811730 

.739961 

1.3514224 

1.24400 

.803857 

30 

40 

.597159 

1.6745970 

.744472 

1.3432331 

1.24669 

.802123 

20 

50 

.599489 

1.6680864 

.749003 

1.3351075 

1.24940 

.800383 

10 

37 

0 

.601815 

1.6616401 

.753554 

1.3270448 

1.25214 

.798636 

0 

53 

10 

.604136 

1.6552575 

.758125 

1.3190441 

1.25489 

.796882 

50 

20 

.606451 

1.6489376 

.762716 

.3111046 

1.25767 

.795121 

40 

30 

.608761 

1.6426796 

.767627 

.3032254 

1.26047 

.793353 

30 

40 

.611067 

1.6364828 

.771959 

.2954057 

1.26330 

.791579 

20 

50 

.613367 

1.6303462 

.776612 

.2876447 

1.26615 

.789798 

10 

38 

0 

.615661 

1.6242692 

.781286 

.2799416 

.26902 

.788011 

0 

52 

10 

.617951 

1.6182510 

.785981 

.2722957 

.27191 

.786217 

50 

20 

.620235 

1.6122908 

.790698 

.2647062 

1.27483 

.784416 

40 

30 

.622515 

1.6063879 

.795436 

.2571723 

1.27778 

.782608 

30 

40 

.624789 

1.6005416 

.800196 

.2496933 

.28075 

.780794 

20 

50 

.627057 

1.5947511 

.804080 

1.2422685 

1.28374 

.778973 

10 

51 

0 

I 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

I 

0 

For  functions  from  51°-10'  to  57°-80'  read  from  bottom  of  table  upward. 

TRIGONOMETRIC  FUNCTIONS 


433 


NATURAL  SINES,  COSINES,  TANGENTS,  ETC. 

(Continued) 

0 

r 

Sine 

Cosecant 

Tangent 

Cotangent 

Secant 

Cosine 

/ 

0 

39 

0 

.629320 

1.5890157 

.809784 

.2348972 

1.28676 

.777146 

0 

51 

10 

.631578 

1.5833318 

.814612 

.2275786 

1.28980 

.775312 

50 

20 

.633831 

1.5777077 

.819463 

.2203121 

1.29287 

.773472 

40 

30 

.636078 

1.5721337 

.824336 

.2130970 

1.29597 

.771625 

30 

40 

.638320 

1.5666121 

.829234 

.2059327 

1.29909 

.769771 

20 

50 

.640557 

1.5611424 

.834155 

.1988184 

1.30223 

.767911 

10 

40 

0 

.642788 

1.5557238 

.839100 

.1917536 

1.30541 

.766044 

0 

50 

10 

.645013 

1.5503558 

.844069 

.1847376 

1.30861 

.764171 

50 

20 

.647233 

1.5450378 

.849062 

1.1777698 

1.31183 

.762292 

40 

30 

.649448 

1.5397690 

.854081 

1.1708496 

1.31509 

.760406 

30 

40 

.651657 

1.5345491 

.859124 

1.1639763 

1.31837 

.758514 

20 

50 

.653861 

1.5293773 

.864193 

1.1571495 

1.32168 

.756615 

10 

41 

0 

.656059 

1.5242531 

.869287 

1.1503684 

1.32501 

.754710 

0 

49 

10 

.658252 

1.5191759 

.874407 

1.1436326 

1.32838 

.752798 

50 

20 

.660439 

1.5141452 

.879553 

1.1369414 

1.33177 

.750880 

40 

30 

.662620 

1.5091605 

.884725 

1.1302944 

1.33519 

.748956 

30 

40 

.664796 

1.5042211 

.889924 

1.1236909 

1.33864 

.747025 

20 

50 

.666966 

1.4993267 

.895151 

1.1171305 

1.34212 

.745088 

10 

42 

0 

.669131 

1.4944765 

.900404 

1.1106125 

1.34563 

.743145 

0 

48 

10 

.671289 

1.4896703 

.905685 

1.1041365 

1.34917 

.741195 

50 

20 

.673443 

1.4849073 

.910994 

1.0977020 

1.35274 

.739239 

40 

30 

.675590 

1.4801872 

.916331 

1.0913085 

1.35634 

.737277 

30 

40 

.677732 

1.4755095 

.921697 

1.0849554 

1.35997 

.735309 

20 

50 

.679868 

1.4708736 

.927091 

1.0786423 

1.36363 

.733335 

10 

43 

0 

.681998 

1.4662792 

.932515 

1.0723687 

1.36733 

.731354 

0 

47 

10 

.684123 

1.4617257 

.937968 

1.0661341 

1.37105 

.729367 

50 

20 

.686242 

1.4572127 

.943451 

1.0599381 

1.37481 

.727374 

40 

30 

.688355 

1.4527397 

.948965 

1.0537801 

1.37860 

.725374 

30 

40 

.690462 

1.4483063 

.954508 

1.0476598 

1.38242 

.723369 

20 

50 

.692563 

1.4439120 

.960083 

1.0415767 

1.38628 

.721357 

10 

44 

0 

.694658 

1.4395565 

.965689 

1.0355303 

1.39016 

.719340 

0 

46 

10 

.696748 

1.4352393 

.971326 

1.0295203 

1.39409 

.717316 

50 

20 

.698832 

1.4309602 

.976996 

1.0235461 

1.39804 

.715286 

40 

30 

.700909 

1.4267182 

.982697 

1.0176074 

1.40203 

.713251 

30 

40 

.702981 

1.4225134 

.988432 

1.0117088 

1.40606 

.711209 

20 

50 

.705047 

1.4183454 

.994199 

1.0058348 

1.41012 

.709161 

10 

45 

0 

.707107 

1.4142136 

1.000000 

1.0000000 

1.41421 

.707107 

0 

45 

o 

I 

Cosine 

Secant 

Cotangent 

Tangent 

Cosecant 

Sine 

t 

o 

For  functions  from  45°  -0'  to  51°-0'  read  from  bottom  of  table  upward. 

APPENDIX  IV 

CONVEKSION   TABLES 


CONVERSION   TABLES 
1 


437 


TABLES  FOR  CONVERTING  UNITED  STATES 

WEIGHTS  AND  MEASURES 

METRIC   TO   CUSTOMARY 

WEIGHTS 

Milligrams 

Crams 

Grams 

Kilograms 

Tonnes  to 

Tonnes  to 

lo. 

to 

to 

to  Avoirdupois 

to  Avoirdupois 

let  Tons  of 

dross  Tons  of 

Grains 

Troy  Ounces 

Ounces 

Pounds 

2000  Founds 

2240  Pounds 

1 

.01543 

.03215 

.03527 

2.20462 

1.10231 

.98421 

2 

.03086 

•06430 

.07055 

4.40924 

2.20462 

1.96841 

3 

.04630 

.09645 

.10582 

6.61387 

3.30693 

2.95262 

4 

.06173 

.12860 

.14110 

8.81849 

4.40924 

3.93682 

5 

.07716 

.16075 

.17637 

11.02311 

5.51156 

4.92103 

(3 

.09259 

.19290 

.21164 

13.22773 

6.61387 

5.90524 

7 

.10803 

.22506 

.24692 

15.43236 

7.71618 

6.88944 

8 

.12346 

.25721 

.28219 

17.63698 

8.81849 

7.87365 

9 

.13889 

.28936 

.31747 

19.84160 

9.92080 

8.85785 

1  Kilogram  =  15432.35639  Grains 

LINEAR   MEASURE 

Millimeters 

Centimeters 

Meters 

Meters 

Kilometers 

Kilometers 

So. 

to  Mths  of  an 

to 

to 

to 

to 

to 

Inch 

Inches 

Feet 

Yards 

Statute  Miles 

Nautical  Miles 

1 

2.51968 

.39370 

3.280833 

1.093611 

.62137 

.53959 

2 

5.03936 

.78740 

6.561667 

2.187222 

1.24274 

1.07919 

3 

7.55904 

1.18110 

9.842500 

3.280833 

1.86411 

1.61878 

4 

10.07872 

1.57480 

13.123333 

4.374444 

2.48548 

2.15837 

5 

12.59840 

1.96850 

16.404167 

5.468056 

3.10685 

2.69796 

6 

15.11808 

2.36220 

19.685000 

6.561667 

3.72822 

3.23756 

7 

17.63776 

2.75590 

22.965833 

7.655278 

4.34959 

3.77715 

8 

20.15744 

3.14960 

26.246667 

8.748889 

4.97096 

4.31674 

9 

22.67712 

3.54330 

29.527500 

9.842500 

5.59233 

4.85633 

438 


CONVERSION   TABLES 
2 


TABLES  FOB  CONVERTING  UNITED  STATES 

WEIGHTS  AND  MEASURES 

CUSTOMARY  TO  METRIC 

WEIGHTS 

drains 

;  Troy  Ounces 

Avoirdupois 

Avoirdupois 

Net  Tons  of 

Gross  Tons  of 

No. 

to 

to 

Ounces 

Pounds  to 

2000  Pounds 

2240  Pounds 

Milligrams 

Grams 

to  Grams 

Kilograms 

to  Tonnes 

to  Tonnes 

1 

64.79892 

31.10348 

28.34953 

.45359 

.90718 

1.01605 

2 

129.59784 

62.20696 

56.69905 

.90718 

1.81437 

2.03209 

3 

194.39675 

93.31044 

85.04858 

1.  36078 

2.72155 

3.04814 

4 

259.19567 

124.41392 

113.39811 

1.81437 

3.62874 

4.06419 

5 

323.99459 

155.51740 

141.74763 

2.26796 

4.53592 

5.08024 

6 

388.79351 

186.62088 

170.09716 

2.72155 

5.44311 

6.09628 

7 

453.59243 

217.72437 

198.44669 

3.17515 

6.35029 

7.11233 

8 

518.39135 

248.82785 

226.79621 

3.62874 

7.25748 

8.12838 

9 

583.19026 

279.93133 

255.14574 

4.08233 

8.16466 

9.14442 

1  Avoirdupois  Pound  =  453.5924277  Grams 

LINEAR  MEASURE 

filths  ..fan 

Inches 

Feet 

Yards 

Statute  Miles 

Nautical  Miles 

No. 

Inch  to 

to 

to 

to 

to 

to 

Millimeters 

Centimeters 

Meters 

Meters 

Kilometers 

Kilometers 

1 

.39688 

2.54001 

.304801 

.914402 

1.60935 

1.85325 

2 

.79375 

5.08001 

.609601 

1.828804 

3.21869 

3.70650 

3 

1.19063 

7.62002 

.914402 

2.743205 

4.82804 

5.5597* 

4 

1.58750 

10.16002 

1.219202 

3.657607 

6.43739 

7.41300 

5 

1.98438 

12.70003 

1.524003 

4.572009 

8.04674 

9.26625 

6 

2.38125 

15.24003 

1.828804 

5.486411 

9.65608 

11.11950 

7 

2.77813 

17.78004 

2.133604 

6.400813 

11.26543 

12.97275 

8 

3.17501 

20.32004 

2.438405 

7.315215 

12.87478 

14.82600 

9 

3.57188 

22.86005 

2.743205 

8.229616 

14.48412 

16.67925 

1  Nautical  Mile    =  1853.25  Meters 

1  Gunter's  Chain  =     20.1168  Meters 

1  Fathom               =       1.829  Meters 

INDEX 


The  numbers  refer  to  the  pages. 


Absorption  dynamometer,    323. 
Acceleration,  184. 

angular,  232. 

components  of,  208. 

constant,  185. 

in  a  curved  path,  205. 

of  points  of  a  body  having  plane 
motion,  237. 

variable,  191. 
Angular  momentum,  369. 

vector  representation  of,  373. 
Angular  velocity,  232,  235. 

vector    representation    of,    241. 
Antifriction  wheels,  298. 
Arch,  linked,  158. 

masonry,  159. 

Balance,  standing  and  running  of  a 

shaft,  359. 
Ball  bearings,  302. 
Bending  moments,  161. 

diagram  of,  162. 

graphical  construction  of,  162. 
Bow's  notation,  99. 
Brake  shoe  testing  machine,  272. 

Car  on  single  rail,  382. 

Catenary,  177. 

Center  of  gravity,  38,  39,  43. 

by  graphical  methods,  57. 

by  integration,  44. 

by  Simpson's  rule,  65. 

motion  of,  377. 

of  counterbalance,  56. 
Center  of  instantaneous  rotation,  405. 
Center  of  oscillation,  345. 
Center  of  percussion,  405. 
Center  of  suspension,  345. 
Coefficient  of  friction,  247,  283. 

of  belting,  316. 
Coefficient  of  restitution,  393. 
Coefficient  of  rolling  friction,  296. 


Compound  pendulum,  344. 
Compression,  14. 
Concurrent  forces,  10,  19. 
Connecting  rod,  366. 
Conservation  of  energy,  254. 
Cords  and  pulleys,  170. 
Couples,  27,  71. 

combination  of,  72,  76. 

components  of,  78. 

equivalent,  73. 

gyroscopic,  380. 

moment  of,  71,  75. 

substitution   of   force   and   couple 
for  a  force,  80. 

vector  representation  of,  77. 
Creeping  of  belts,  315. 

D'Alembert's  principle,  334. 
Determination  of  g,  347. 
Displacement,  3. 
Durand's  rule,  68. 
Dynamometer,  absorption,  323. 
transmission,  314. 

Effective  forces,  267,  334. 
Efficiency  of  wedge,  288. 
Elasticity  of  material,  397. 

modulus  of,  398. 
Ellipse  of  inertia,  130. 
Ellipsoid  of  inertia,  140. 
Energy,  254. 
Equilibrant,  94. 

graphical  method  of  finding,  94. 
Equilibrium,  1. 

conditions  of,  12,  20,  29,  83,  92. 

of  three  forces,  16. 

of  two  forces,  14. 

Falling  bodies,  186. 
Flexible  cords,  148. 

as  a  catenary,  177. 

as  a  parabola,  171. 


439 


440 


INDEX 


Force,  1. 

transmissibility  of,  6. 

unit  of,  2. 

vector  representation  of,  5. 
Forces,  concurrent,  6,  10,  19. 

in  space,  90. 

non-concurrent,  82. 

polygon  of,  8. 

resolution  of,  7. 

triangle  of,  7. 
Foucault's  pendulum,  243. 
Friction  brake,  324. 

Prony,  326. 
Friction  circle,  317. 
Friction,  coefficient  of,  247,  283. 

laws  of,  for  dry  surfaces,  285. 

of  belts,  310. 

of  brake  shoe,  328. 

of  lubricated  surfaces,  288. 

of  pivots,  318. 

of  worn  bearing,  316. 

rolling,  296. 
Friction  gears,  307. 

Guldinus,  theorem  of,  68. 
Gyroscope,  379. 

inclined  axis,  383. 
Gyroscopic  couple,  380. 

Harmonic  motion,  192. 
Helical  chute,  229. 
Horse  power,  253. 
Hyperbolic  functions,  180. 
tables  of,  appendix. 

Impact,  direct,  oblique,  central,  390. 

direct  central,  391. 

direct  eccentric,  401. 

oblique  against  smooth  plane,  408. 

of  rotating  bodies,  411. 

on  a  body  with  fixed  axis,  403. 
Impact  tension  and  compression,  398. 
Impressed  forces,  266,  334. 
Impulse  of  a  force,  394. 
Inertia,  1. 

ellipse  of,  130. 

ellipsoid  of,  140. 

moment  of,  106. 

product  of,  112. 
Instantaneous  axis  of  rotation,  238, 407. 


center  of  rotation,  239,  407. 
Internal  stress  couple,  166. 

Kinetic  energy,  254. 
lost  in  impact,  395. 
of  a  body  having  plane  motion,  274. 
of  a  body  with  one  fixed  point,  373. 
of  rolling  bodies,  279. 

Line  of  resistance  of  arch,  159. 
Lubricants,  testing  of,  293. 

Mass,  1. 

unit  of,  2,  188. 

engineer's  unit  of,  188. 
Method  of  substitution,  103. 
Modulus  of  elasticity,  167,  398. 
Moment  of  a  force,  22,  24. 

of  the  resultant,  23,  25,  28. 
Moment  of  inertia,  106,  268. 

axes  of  greatest  and  least,  112,  128. 

by  direct  addition,  126. 

by  experiment,  346,  351. 

by  graphical  method,  124. 

by  parallel  sections,  135. 

by  Simpson's  rule,  126. 

polar,  115. 

principal  moments  of  inertia,  139. 

theorem  of  inclined  axes  for,  111. 

theorem  of  parallel  axes  for,  109. 

units  of,  108. 
Moment  of  momentum,  369. 

of  a  body  with  one  fixed  point,  371. 

rate  of  change  of,  due  to  rotating 

axes,  375. 

Momentum  of  a  body,  394. 
Motion,  along  a  curve  in  a  vertical 
plane,  211. 

curvilinear,  204. 

in  a  circle,  233. 

in  a  twisted  curve,  227. 

of  center  of  gravity  of  body,  377. 

on  inclined  plane,  189. 

rotary,  232. 

uniform,  in  a  circle,  209. 

where  attraction  varies  directly  as 
distance  squared,  197. 

where  resistance  varies  as  distance, 
194. 

with  repulsive  force,  194. 


INDEX 


441 


Neutral  surface,  168. 
Newton's  laws  of  motion,  187. 

Pappus,  theorem  of,  68. 

Parabola,  171. 

Parallel  forces,  center  of,  32,  35. 

resultant  of,  27,  29. 
Pendulum,  constant  of,  398. 

compound,  344. 

conical,  209. 

cycloidal,  219. 

simple,  214. 
Period,  of  compound  pendulum,  345. 

of  deformation,  391. 

of  restitution,  391. 

of  simple  pendulum,  216. 
Piledriver,  259. 
Pivots,  flat  end,  318. 

collar  bearing,  318. 

conical,  319. 

Schiele's,  322. 

spherical,  321. 

Polar  moment  of  inertia,  115. 
Pole  of  string  polygon,  154. 
Power,  transmitted  by  a  belt,  313. 

transmitted  by  friction  wheel,  309. 
Precession,  380. 

steady,  384. 

unsteady,  386. 
Principal  axes,  113,  139. 
Principal  moments  of  inertia,  139. 
Product  of  inertia,  112. 
Projectile  in  vacuo,  222. 

in  resisting  medium,  226. 

Radius  of  gyration,  107. 

Reaction    of    supports    of    rotating 

body,  340. 

Relative  velocity,  200. 
Resistance  of  roads,  299. 
Rigid  body,  4. 
Roller  bearings,  301. 
Rolling  friction,  296. 
Rotation  about  a  fixed  axis,  266,  336 

with  constant  angular  velocity,  353 
Rotation  and  translation,  365. 
Rotation,  of  flywheel  of   steam    en 
gine,  362. 

of  locomotive  drive  wheel,  358. 


of  the  earth,  243. 
under  no  forces,  352. 

Shear,  166. 
Simpson's  rule,  62. 

applications  of,  64. 
Simultaneous    rotation    about   inter- 
secting axes,  240. 
pecific  gravity,  table  of,  3. 
peed,  183. 

Steam  hammer,  262. 
Stresses  in  beams,  166. 
Stresses  in  frames,  99. 
String  polygon,  153. 
depth  of,  156. 
locus  of  pole  of,  154. 
Substitution,  method  of,  103. 

Tension,  14. 

Torque  and  angular  momentum,  370. 

Torsion  balance,  348. 

Train  resistance,  330. 

Translation  of  rigid  body,  335. 

Transmission  dynamometer,  314. 

Unit  strain,  167. 
Unit  stress,  167. 
Unit  weights,  table  of,  3. 

Varignon's  theorem,  23. 
Vector  rate  of  change  due  to  rota- 
tion, 374. 
Vector     representation     of     angular 

momentum,  373. 
Vectors,  3. 
Velocity,  183. 
of  spin,  383. 
relative,  200. 

Weight,  3. 

unit  of,  3. 
Weights  suspended  by  cords,  150. 

graphical  solution  of,  151. 
Work  and  energy,  245,  255,  276,  279. 

units  of,  248. 
Work  of  a  variable  force,   250. 

of  components  of  a  force,  249. 

graphical  representation  of,  251. 

in  uniform  motion,  280. 


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